Applying L'Hôpital's rule infinitely I tried to prove that $\int\limits_0^\infty t^{x-1} e^{-t} \, \mathrm{d}t$ satisfies the functional equation of the gamma function $\Gamma(x+1)=x\Gamma(x)$, so I partially integrated $\Gamma(x+1)$, yielding $\left[-e^{-t}\,t^x\right]_0^\infty+x \Gamma(x)$.
It is obvious to me that
$$\lim_{a\to\infty} \frac{a^x}{e^a}$$ is zero for any finite $a$ due to the fact that the exponential grows faster than any polynomial. You could show this quite easily by using L'Hôpital's rule $a$ times. I can imagine, however, that this is not true in the non-finite case. Can I apply L'Hôpital in this way, what would I have to show in order to do so, and if I cannot, please give me a hint how to obtain the desired result in a different way.

Basically, there is no infinite case. $\int\limits_0^\infty t^{x-1} e^{-x} \, \mathrm{d}t$ is defined to be $\lim_{N \rightarrow \infty} \int\limits_0^N t^{x-1} e^{-x} \, \mathrm{d}t$.
So the way to prove such identities is to fix $N$ and then integrate by parts in $\int\limits_0^N t^{x-1} e^{-x} \, \mathrm{d}t$ as you have done here (more than once if needed). Then take limits of the resulting identity as $N$ goes to infinity. In your example you'll end out using L'hopital's rule as you've done to show the boundary terms go to zero.
