Let $z\in\mathbb{C}$. We know that $$\dfrac{1}{\Gamma(z)}=e^{\gamma z}\prod_{n=1}^\infty\left(1+\dfrac{z}{n}\right)e^{-z/n},$$ where $$\gamma=\lim_{N\rightarrow\infty}\sum_{n=1}^N\dfrac1n-\log N.$$
How can we use this to prove that $$\Gamma(z)=\lim_{n\rightarrow\infty}\dfrac{n^zn!}{z(z+1)\ldots(z+n)}?$$
Write $\Gamma(z)$ as the limit of the partial products,
$$\begin{align} \Gamma(z) &= \lim_{n\to \infty} e^{-\gamma z}\frac1z\prod_{k=1}^n \left(1+\frac{z}{k}\right)^{-1}e^{z/k}\\ &= \lim_{n\to\infty} \frac{n!}{\prod_{k=0}^n (z+k)}\exp \left(z\left(\sum_{k=1}^n \frac1k - \gamma\right)\right)\\ &= \lim_{n\to\infty} \frac{n! n^z}{z(z+1)\dotsb(z+n)}\exp\left(z\left(\sum_{k=1}^n \frac1k - \log n - \gamma\right)\right). \end{align}$$
Since $\sum\limits_{k=1}^n \frac1k - \log n - \gamma \xrightarrow{n\to\infty} 0$, the second factor converges to $1$, hence
$$\Gamma(z) = \lim_{n\to\infty} \frac{n! n^z}{z(z+1)\dotsb (z+n)}.$$