For $x\in (0,\infty)$, show that $\underset{n\to\infty}{\text{lim}}\frac{n^xn!}{x^{<n+1>}}=\Gamma(x)$.
Hints: Use Stirling's formula: $\underset{x\to\infty}{\text{lim}}\frac{\Gamma(x+1)}{(\frac{x}{e})^x\sqrt{2\pi x}}=1$. Also, $\Gamma(x+n+1)=x^{<n+1>}\Gamma(x)$, where $x^{<x+1>}=x(x+1)...(x+n-1)(x+n)$.
That is usually done by noticing that $$ \frac{n^x n!}{x^{\langle n+1\rangle}}=\int_{0}^{n} z^{x-1}\left(1-\frac{z}{n}\right)^{n}\,dz \tag{1}$$ holds by integration by parts, then by applying the dominated/monotone convergence theorem to prove that the limit of the RHS of $(1)$ as $n\to +\infty$ is $$ \int_{0}^{+\infty} z^{x-1} e^{-z}\,dz = \Gamma(x).\tag{2}$$ All of that before proving Stirling's inequality, usually.
With $\beta$ function here we know $\beta(x,y)=\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ and fixed $x$ with large $n$ we have $\beta(x,n)=\Gamma(x)n^{-x}$ then $$\lim_{n\to\infty}\dfrac{n^xn!}{x^{<n+1>}}=\lim_{n\to\infty}\dfrac{n^x\Gamma(n+1)\Gamma(x)}{\Gamma(x+n+1)}=\lim_{n\to\infty}n^x\beta(n+1,x)=\lim_{n\to\infty}n^x\Gamma(x)(n+1)^{-x}=\Gamma(x)$$
To expand on the answer left by @JackD'Aurizio, we let $G_n(x)$ be the sequence of functions given by
$$G_n(x)=\int_0^n s^{x-1}\left(1-\frac{s}{n}\right)^n\,ds$$
We will show without using the Dominated Convergence Theorem that $\lim_{n\to \infty}G_n(x)=\Gamma(x)$ by showing that
$$\lim_{n\to \infty}\int_0^n s^{x-1}e^{-s}\left(1-e^s\left(1-\frac{s}{n}\right)^n\right)=0 \tag 1$$
To do this, we appeal to the analysis in THIS ANSWER to obtain bounds for the integrand in $(1)$. Proceeding, we have
$$\begin{align} 1-e^s\left(1-\frac{s}{n}\right)^n &\le 1-\left(1+\frac{s}{n}\right)^n\left(1-\frac{s}{n}\right)^n\\\\ &=1-\left(1-\frac{s^2}{n^2}\right)^n\\\\ &\le 1-\left(1-\frac{s^2}{n}\right)\\\\ &=\frac{s^2}{n} \end{align}$$
where Bernoulli's Inequality was used to arrive at the last inequality. Similarly, we see that
$$\begin{align} 1-e^s\left(1-\frac{s}{n}\right)^n &\ge 1-e^se^{-s}\\\\ &=0 \end{align}$$
Therefore, applying the squeeze theorem yields to coveted limit
$$\lim_{n\to \infty}\int_0^n s^{x-1}e^{-s}\left(1-e^s\left(1-\frac{s}{n}\right)^n\right)=0$$
which implies $\lim_{n\to \infty}G_n(x)=\Gamma(x)$.
Integrating by parts repeatedly the integral representation of $G_n(x)$ reveals
$$G_n(x)=\frac{n^x\,n!}{x(x+1)(x+2)\cdots (x+n)}$$
so that
$$\bbox[5px,border:2px solid #C0A000]{\Gamma(x)=\lim_{n\to \infty}\frac{n^x\,n!}{x(x+1)(x+2)\cdots (x+n)}}$$
And we are done.
