Why this gamma function reduces to the factorial? $$\Gamma(m+1) = \frac{1\cdot2^m}{1+m}\frac{2^{1-m}\cdot3^m}{2+m}\frac{3^{1-m}\cdot4^m}{3+m}\frac{4^{1-m}\cdot5^m}{4+m}\cdots$$
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.
