Proof involving gamma function, infinite product and Gauss

Question

How can I rigorously and directly prove that $$\Gamma (z)=\lim_{n\rightarrow \infty }\frac{n!n^{z}}{z(z+1)\cdots(z+n)}$$

Answer 1

For simplicity define $z=k$ ,so we prove

$$\lim_{n\to \infty}\frac{n! \,\, n^k}{k(k+1)\cdots (k+n)}=(k-1)!$$

Note that

$$\frac{(k+n)!}{(k-1)!} = k(k+1)\cdots (k+n)$$

so we have

$$ (k-1)! \lim_{n\to \infty}\frac{ \, n! \,\, n^k}{(k+n)!}=(k-1)!\lim_{n\to \infty}\frac{n^k}{(k+n)(k+n-1)\cdots (n+1)}$$

Now this can be written as

$$(k-1)! \lim_{n\to \infty} \frac{n\cdot n\cdot \cdots n}{(n+k)(n+k-1)\cdots (n+1)}=(k-1)!$$

The gamma function is an extension of the factorial by the relation

$$\Gamma(k)=(k-1)!$$