I came across a proof of Gauss multiplication formula for the Gamma function which relies on the following indentity (without a proof)
$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=m^{mn}\prod_{k=1}^{m}\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)} \, \tag{1}$$
I am trying to prove it.
I started expanding the left hand side first. From the recurrence relation of the Gamma function we have that
$$\Gamma \left(x+mn \right)= \left(x+mn -1\right)\Gamma \left(x+mn-1 \right)$$
$$\Gamma \left(x+mn \right)= \left(x+mn -1\right) \left(x+mn -2\right)\Gamma \left(x+mn-2 \right)$$
$$\cdots$$
$$\Gamma \left(x+mn \right)= \left(x+mn -1\right) \left(x+mn -2\right) \cdots \left(x+mn -mn\right)\Gamma \left(x+mn-mn \right)$$
Therefore we can rewrite the L.H.S of $(1)$ as
$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=\left(x+mn -1\right) \left(x+mn -2\right) \cdots \left(x\right)$$
$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=\prod_{r=1}^{mn}\left(x+mn -r\right) \, \tag{2}$$
Similarly, for the R.H.S. of $(1)$ we obtain
$$\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)}=\left(\frac{x+k-1}{m}+n-1 \right)\left(\frac{x+k-1}{m}+n-2 \right) \cdots \left(\frac{x+k-1}{m}+n-n \right)$$
$$\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)}=\prod_{l=1}^{n}\left(\frac{x+k-1}{m}+n-l \right)$$
$$\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)}=\prod_{l=1}^{n}\frac{1}{m}\left(x+k-1+mn-ml \right) \, \tag{3}$$
Plugging $(2)$ and $(3)$ in $(1)$ we obtain the following equality.
$$\prod_{r=1}^{mn}\left(x+mn -r\right)=\prod_{k=1}^{m}\prod_{l=1}^{n}\left(x+k-1+mn-ml \right) \, \tag{4} $$
$$\prod_{r=1}^{mn}\left(x+mn -r\right)=\prod_{k=1}^{m}\prod_{l=1}^{n}\left(x+mn-(1+ml-k) \right)$$
suposse $m=2 \,\, \text{and}\,\,n=2$, the right hand side becomes
$$\prod_{k=1}^{2}\prod_{l=1}^{2}\left(x+k-1+4-2l \right)=\prod_{k=1}^{2}\left(x+k-1+4-2 \right)\left(x+k-1+4-2\times2 \right)$$
$$=\prod_{k=1}^{2}\left(x+k+1 \right)\left(x+k-1 \right)$$
$$=\left(x+1+1 \right) \cdot\left(x+1-1 \right)\cdot\left(x+2+1 \right)\cdot\left(x+2-1 \right)$$
$$=\left(x+2 \right)\cdot x \cdot \left(x+3 \right) \cdot\left(x+1 \right)$$
$$= x \cdot\left(x+1 \right)\cdot \left(x+2 \right)\cdot \left(x+3 \right) \, \tag{5}$$
And the left hand side becomes
$$\prod_{r=1}^{4}\left(x+4 -r\right)=\left(x+4 -1\right)\left(x+4 -2\right)\left(x+4 -3\right)\left(x+4 -4\right)$$
$$= x \cdot\left(x+1 \right)\cdot \left(x+2 \right)\cdot \left(x+3 \right)$$
Which equals exactly $(5)$. So intuitively I see that the equality $(1)$ holds. My question is: How can I go from this heuristic intuitive proof to a formal proof, may be proved by induction?
Ok, after a long arduous work I got it. Consider $(1)$ above
$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=m^{mn}\prod_{k=1}^{m}\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)} \, \tag{1}$$
Note that both sides of (1), the ration between Gamma functions, can be rewritten as Pocchammer symbols. By the recurrence equation of the Gamma function, namely
$$\Gamma \left(x+1 \right)=x\Gamma \left(x+1 \right)$$
we may write
$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=\frac{\left(x+mn-1 \right)\Gamma \left(x+mn-1 \right)}{\Gamma \left(x \right)}$$
$$\cdots$$
$$=\frac{\left(x+mn-1 \right)\left(x+mn-2 \right) \cdots x\Gamma \left(x \right)}{\Gamma \left(x \right)}$$
$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=x (x+1) \cdots\left(x+mn-2 \right)\left(x+mn-1 \right) $$
$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=\left( x\right)_{mn}\, \tag{2}$$
enforcing $x \longrightarrow \frac{x+k-1}{m} \, \, \text{and}\,\, mn \longrightarrow n$ in $(2)$ above we get
$$\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)}=\left(\frac{x+k-1}{m} \right)_{n}\, \tag{3}$$
plugging $(2)$ and $(3)$ in $(1)$ we get
$$\left( x\right)_{mn}=m^{mn}\prod_{k=1}^{m}\left(\frac{x+k-1}{m} \right)_{n}\tag{4}$$
Now, lets play a little with Pocchammer symbol to see whether we can draw something from it. By definition we have
$$\left( x\right)_{m}=\underbrace{x(x+1)(x+2) \cdots (x+m-2)(x+m-1)}_{\text{n terms}}$$
We have one group of m terms each
$$\left( x\right)_{2m}=\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)}\color{blue}{(x+m-1+1)(x+m-1+2)\cdots(x+m-1+m)}$$
$$\left( x\right)_{2m}=\underbrace{\underbrace{\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)}}_{\text{n terms}}\underbrace{\color{blue}{(x+m)(x+m+1)\cdots(x+m+m-1)}}_{\text{n terms}}}_{\text{2n terms}}$$
we have 2 groups of m terms each
$$\left( x\right)_{3m}=\underbrace{\underbrace{\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)}}_{\text{n terms}}\underbrace{\color{blue}{(x+m)(x+m+1)\cdots(x+m+m-1)}}_{\text{n terms}}\color{green} {\underbrace{(x+2m)(x+2m+1) \cdots(x+2m+m-1)}_{\text{n terms}}}}_{\text{3n terms}}$$
We have 3 groups of m terms each
Generalizing we may write
$$\left( x\right)_{nm}=\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)} \color{black}{\cdots} \color{blue}{(x+(n-1)m)(x+(n-1)m+1)\cdots(x+(n-1)m+m-1)}$$
$$\left( x\right)_{nm}=\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)} \color{black}{\cdots} \color{blue}{(x+(n-1)m)(x+(n-1)m+1)\cdots(x+nm-1)}$$
We obtain n groups of m terms each.
Now, instead o writing each group horizontally, lets write each group in a different line getting sort of a matrix look. For $(x)_{2m}$
$$ \begin{aligned} &\begin{array}{llcclll} (x)_{2m}=&x & (x+1) & (x+2) & \cdots & (x+m-1) \\ &\left(x+m\right) & \left(x+m+1\right) & \left(x+m+2\right) & \ldots & \left(x+m+m-1 \right) \\ \end{array}\\ \end{aligned} $$
$$ \begin{aligned} &\begin{array}{llcclll} (x)_{2m}=&m^m& \frac{x}{m} & (\frac{x+1}{m}) & (\frac{x+2}{m}) & \cdots & (\frac{x+m-1)}{m} \\ &m^m&\left(\frac{x}{m}+1\right) & \left(\frac{x+1}{m}+1\right) & \left(\frac{x+2}{m}+1\right) & \ldots & \left(\frac{x+m-1}{m}+1 \right) \end{array}\\ \end{aligned} $$
Now multiply vertically along each column to obtain
$$(x)_{2m}=m^{2m} \cdot \left(\frac{x}{m}\right)_{2} \cdot\left(\frac{x+1}{m}\right)_{2} \cdot \left(\frac{x+2}{m}\right)_{2} \cdots \left(\frac{x+m-1}{m}\right)_{2}$$
And we may write
$$(x)_{2m}=m^{2m}\prod_{k=0}^{m-1}\left(\frac{x+k}{m} \right)_{2}$$
Similarly
$$ \begin{aligned} &\begin{array}{llcclll} (x)_{3m}=&x & (x+1) & (x+2) & \cdots & (x+m-1) \\ &\left(x+m\right) & \left(x+m+1\right) & \left(x+m+2\right) & \ldots & \left(x+m+m-1 \right) \\ &\left(x+2m\right) & \left(x+2m+1\right) & \left(x+2m+2\right) & \ldots & \left(x+2m+m-1 \right) \end{array}\\ \end{aligned} $$
$$ \begin{aligned} &\begin{array}{llcclll} (x)_{3m}=&m^m& \frac{x}{m} & (\frac{x+1}{m}) & (\frac{x+2}{m}) & \cdots & (\frac{x+m-1)}{m} \\ &m^m&\left(\frac{x}{m}+1\right) & \left(\frac{x+1}{m}+1\right) & \left(\frac{x+2}{m}+1\right) & \ldots & \left(\frac{x+m-1}{m}+1 \right) \\ &m^m&\left(\frac{x}{m}+2\right) & \left(\frac{x+1}{m}+2\right) & \left(\frac{x+2}{m}+2\right) & \ldots & \left(\frac{x+m-1}{m}+2 \right) \\ (x)_{3m}=&m^{3m}& \left(\frac{x}{m}\right)_{3}&\left(\frac{x+1}{m}\right)_{3}&\left(\frac{x+2}{m}\right)_{3}& \cdots &\left(\frac{x+m-1}{m}\right)_{3} \end{array}\\ \end{aligned} $$
And we may write
$$(x)_{3m}=m^{3m}\prod_{k=0}^{m-1}\left(\frac{x+k}{m} \right)_{3}$$
Generalizing, we may write
$$(x)_{nm}=A_{1} \cdot A_{2} \cdots A_{n} $$
where
$$ \begin{array}{cccccc} A_{1}= & x & (x+1) & (x+2) & \ldots & (x+m-1) \\ A_{2}= & (x+m) & (x+m+1) & (x+m+2) & \ldots & (x+m+ m-1) \\ A_{3}= & (x+2 m) & (x+2 m+1) & (x+2 m+2) & \ldots & (x+2m+ m-1) \\ \vdots & \vdots & \vdots & \ldots & \ldots & \vdots \\ A_{n}= & (x+(n-1) m) & (x+(n-1) m+1) & (x+(n-1) m+2) & \ldots & (x+(n-1)m+ m-1) \end{array} $$
Factoring out m in each line
$$ \begin{aligned} &\begin{array}{llcclll} A_{1}= & m^{m} & \frac{x}{m} & \left(\frac{x+1}{m}\right) & \left(\frac{x+2}{m}\right) & \ldots & \left(\frac{x+m-1}{m}\right) \\ A_{2}= & m^{m} & \left(\frac{x}{m}+1\right) & \left(\frac{x+1}{m}+1\right) & \left(\frac{x+2}{m}+1\right) & \ldots & \left(\frac{x+m-1}{m}+1\right) \\ A_{3}= & m^{m} & \left(\frac{x}{m}+2\right) & \left(\frac{x+1}{m}+2\right) & \left(\frac{x+2}{m}+2\right) & \ldots & \left(\frac{x+ m-1}{m}+2\right)\\ \vdots & \vdots & \vdots & \ldots & \ldots & \vdots\\ A_{n}=&m^{m} &\left(\frac{x}{m}+n-1\right)& \left(\frac{x+1}{m}+n-1\right) &\left(\frac{x+2}{m}+n-1\right)& \ldots &\left(\frac{x+m -1}{m}+n-1\right)\\ \end{array}\\ \end{aligned} $$
Which analogously as the above cases, we may then write
$$(x)_{nm}=m^{nm}\left(\frac{x}{m} \right)_{n}\left(\frac{x+1}{m} \right)_{n} \cdots \left(\frac{x+m-1}{m} \right)_{n}$$
and finally
$$(x)_{nm}=m^{nm}\prod_{k=0}^{m-1}\left(\frac{x+k}{m} \right)_{n}$$
or shifting the index we get exactly $(4)$
$$(x)_{nm}=m^{nm}\prod_{k=1}^{m}\left(\frac{x+k-1}{m} \right)_{n}$$