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:
$$\frac{1\cdot2\cdot3\cdots n\cdot(n+1)^m}{(1+m)(2+m)\cdots(n+m)}$$

This involves the fact that $\Gamma(1)=1$ and $\Gamma(n+1)=n\Gamma(n)$ based on the definition of Gamma, $\Gamma(t)=\int_0^\infty x^{t-1}e^{-x}dx$.
