I don't have a copy of Rudin, but can give you a proof of this. There's various ways of playing about with the integral form to get what you want, although I'm not sure what the cleanest method is.
One way is to use the fact that $1_{\{t\le n\}}(1-t/n)^n\to e^{-t}$ to write $\Gamma(x)$ as
$$
\begin{align}
\Gamma(x)&=\lim_{n\to\infty}\int_0^nt^{x-1}(1-t/n)^{n}\,dt\\
&=\lim_{n\to\infty}n^x\int_0^1s^{x-1}(1-s)^n\,ds.
\end{align}
$$
To prove that the first integral commutes with the limit, you could use the dominated convergence theorem. The second integral is just using the substitution $t=ns$.
It needs to be shown that this integral is equal to your function $f_n(x)$. In fact, I recognize both the function $f_n$ and the integral as beta functions.
$$
\int_0^1s^{x-1}(1-s)^n\,ds=B(x,n+1)
$$
To find the explicit form for this, you can either use $B(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)$ (a proof is given on the Wikipedia page) or repeatedly apply the identity $B(x,n+1)=B(x+1,n)n/x$ (which follows from integration by parts) and $B(x,1)=1/x$.
Note: You can also apply a very similar argument to the one above using the limit $(1+t/n)^{-n}\to e^{-t}$, which I did in my first version of this answer. Using $(1-t/n)^n\to e^{-t}$ seems cleaner though, so I edited the answer accordingly.
An alternative method is to write
$$
\begin{align}
\Gamma(x) &= \frac{\Gamma(x+n+1)}{\Gamma(n+1)}\frac{\Gamma(n+1)\Gamma(x)}{\Gamma(x+n+1)}\\
&= \frac{\Gamma(x+n+1)}{\Gamma(n+1)}\frac{n!}{x(x+1)\cdots(x+n)}.
\end{align}
$$
Then, the limit you need would follow as long as it can be shown that $\Gamma(x+n+1)/\Gamma(n+1)$ approaches $n^x$ asymptotically as $n\to\infty$.
$$
\Gamma(n+x+1)=\int_0^\infty t^x t^ne^{-t}\,dt
$$
Then, the required expression depends on showing that, in the limit $n\to\infty$, to leading order, the integral only contributes for values of $t/n$ close to 1 (you can see that $t^ne^{-t}$ has its maximum at $t=n$).