$$\Gamma(m+1) = \frac{1\cdot2^m}{1+m}\frac{2^{1-m}\cdot3^m}{2+m}\frac{3^{1-m}\cdot4^m}{3+m}\frac{4^{1-m}\cdot5^m}{4+m}\cdots$$
My books says that in a letter from Euler to Goldbach, this expression reduces to $m!$ when $m$ is a positive integer, but that Euler verified it only for $m=2$ and $m=3$
How can I verify it?
Also, the book shows this other form:
$$\frac{1\cdot2\cdot3\cdots n\cdot(n+1)^m}{(1+m)(2+m)\cdots(n+m)}$$
The Product $$ \begin{align} \prod_{k=1}^n\frac{\color{#C00000}{k^{1-m}}\color{#00A000}{(k+1)^m}}{\color{#0000FF}{k+m}} &=\color{#C00000}{n!^{1-m}}\color{#00A000}{(n+1)!^m}\color{#0000FF}{\frac{m!}{(n+m)!}}\\ &=n!\left(\frac{(n+1)!}{n!}\right)^m\frac{m!}{(n+m)!}\\ &=n!(n+1)^m\frac{m!}{(n+m)!}\\ &=m!\color{#00A000}{(n+1)^m}\color{#0000FF}{\frac{n!}{(n+m)!}}\\ &=m!\prod_{k=1}^m\frac{\color{#00A000}{n+1}}{\color{#0000FF}{n+k}}\\ &=m!\prod_{k=1}^m\frac{1+\frac1n}{1+\frac kn}\tag{1} \end{align} $$ Taking the limit as $n\to\infty$ yields $$ \prod_{k=1}^\infty\frac{k^{1-m}(k+1)^m}{k+m}=m!\tag{2} $$
Comment 1
Bernoulli's Inequality says that $$ 1+\frac mk\le\left(1+\frac1k\right)^m\tag{3} $$ and each term in the product of $(2)$ is $$ \begin{align} \frac{k^{1-m}(k+1)^m}{k+m} &=\frac{\left(1+\frac1k\right)^m}{1+\frac mk}\\ &\ge1\tag{4} \end{align} $$ Thus, $(4)$ says that each term of the product is at least $1$.
Comment 2
The Binomial Theorem says that for $k\ge1$, $$ \begin{align} \left(1+\frac1k\right)^m &=1+\frac mk+\sum_{j=2}^m\binom{m}{j}\frac1{k^j}\\ &\le1+\frac mk+\frac{2^m}{k^2}\tag{5} \end{align} $$ Therefore, each term in the product $(2)$ is $$ \begin{align} \frac{k^{1-m}(k+1)^m}{k+m} &=\frac{\left(1+\frac1k\right)^m}{1+\frac mk}\\ &\le1+\frac{2^m}{k^2}\\ &\le e^{2^m/k^2}\tag{6} \end{align} $$ Taking the product of $(6)$ gives $$ \prod_{k=1}^\infty\frac{k^{1-m}(k+1)^m}{k+m}\le e^{2^m\pi^2/6}\tag{7} $$ Thus, $(7)$ says that the product is finite.