Question: How to prove the following identity for all positive integers $k$ and $n$: \begin{align}\tag{1} (k+1)(2k+1) \cdots (nk+1) = \sum_{i=1}^n & \binom{n}{i} \left( \frac{i}{n}(k+2)-1 \right) \left[ (k+1)(2k+1) \cdots ((i-1)k+1) \right] \\ &\ \times \left[ (k+1)(2k+1) \cdots ((n-i)k+1) \right]. \end{align}
I could check by enumeration in $n$:
\begin{align} n&=1, & k+1 &= \binom{1}{1} (k+2-1), \\ n&=2, & (k+1)(2k+1) &= \binom{2}{1} \left( \frac{1}{2}(k+2)-1 \right) (k+1) + \binom{2}{2} \left( (k+2)-1 \right) (k+1) \\ & & & = k(k+1) + (k+1)^2, \\ n&=3, & (k+1)(2k+1)(3k+1) &= \binom{3}{1} \left( \frac{1}{3}(k+2)-1 \right) (k+1)(2k+1) \\ & & &\quad + \binom{3}{2} \left( \frac{2}{3}(k+2)-1 \right) (k+1)^2 \\ & & &\quad + \binom{3}{3} \left( (k+2)-1 \right) (k+1)(2k+1) \\ & & & = (k-1)(k+1)(2k+1) + (2k+1)(k+1)^2 + (k+1)^2(2k+1), \\ & \cdots & & \end{align} But I failed to prove it by induction in $n$: Set \begin{equation} I(n;k) := \sum_{i=1}^n \binom{n}{i} \left( \frac{i}{n}(k+2)-1 \right) \left[ (k+1) \cdots ((i-1)k+1) \right] \left[ (k+1) \cdots ((n-i)k+1) \right]. \end{equation} Then \begin{equation} \begin{split} I(n+1;k) - I(n;k) =&\ (k+1) \left[ (k+1) \cdots (nk+1) \right] \\ &\ + \sum_{i=1}^n \binom{n}{i} \left[ (k+1) \cdots ((i-1)k+1) \right] \left[ (k+1) \cdots ((n-i)k+1) \right] \\ &\ \qquad\quad \times \underbrace{\left[ \frac{n+1}{n+1-i} \left( \frac{i}{n+1}(k+2)-1 \right) ((n+1-i)k+1) - \left( \frac{i}{n}(k+2)-1 \right) \right]}_{=:J(n;k,i)} \end{split} \end{equation} If the followin equation holds, \begin{equation}\tag{2} J(n;k,i) = \left( \frac{i}{n}(k+2)-1 \right)(nk-1), \end{equation} then using the inductive hypothesis $I(n;k) = (k+1) \cdots (nk+1)$, we have \begin{equation} \begin{split} I(n+1;k) - I(n;k) =&\ (k+1) \left[ (k+1) \cdots (nk+1) \right] + (nk-1) I(n;k) = (n+1)k I(n;k) \end{split} \end{equation} which yields $I(n+1;k) = ((n+1)k+1) I(n;k) = (k+1) \cdots (nk+1) ((n+1)k+1)$ as desired.
But, unfortunately, equation $(2)$ does not hold...
I was also thinking that the question may be related to Gamma function, because $$ (k+1)(2k+1) \cdots (nk+1) = k^n \frac{\Gamma(n+1+\frac{1}{k})}{\Gamma(1+\frac{1}{k})} $$ so that identity $(1)$ turns to \begin{equation}\tag{1'} k \Gamma\left(n+1+\frac{1}{k}\right) \Gamma\left(1+\frac{1}{k}\right) = \sum_{i=1}^n \binom{n}{i} \left( \frac{i}{n}(k+2)-1 \right) \Gamma\left(i+\frac{1}{k}\right) \Gamma\left(n-i+1+\frac{1}{k}\right). \end{equation} But still, I have no clue to prove $(1')$.
Could anyone help on it? Any hint or comment will be appreciated. TIA...
Let $$A_m=(k+1)(2k+1)...(mk+1)$$
Let $$B_{n+1}=\sum_{i=1}^{n+1}\left[(k+2){n\choose i-1}-{n+1\choose i}\right]A_{i-1}A_{n+1-i}$$
Break both the binomial coefficients according to Pascal's rule. The terms with $(k+2)$ factors can then be paired up again, giving terms of the form $${n-1\choose i-1}(A_{i-1}A_{n+1-i}+A_iA_{n-i})\\ ={n-1\choose i-1}A_{i-1}A_{n-i}((n+1)k+2)$$
The other terms can be combined similarly, but there is an $A_n$ left over from the $i=1$ term. From comparison with $B_n$, we get $$B_{n+1}=((n+1)k+2)B_n-A_n$$ So, by induction $B_n=A_n$