# Showing that $\lim_{t \rightarrow \infty}-e^{-t}t^x=0$

I'm trying to prove that $\Gamma(x+1) = x \Gamma(x)$ and after doing integration by parts on $\Gamma(x+1)$ I'm left with a term $-t^xe^{-t}$ that I need to evaluate at the limit as $t \rightarrow \infty$.

I was thinking about simply stating that by applying L'Hospital $\left\lceil x \right\rceil$-times would give: $$\displaystyle 0 \leq \lim_{t \rightarrow \infty} \frac{-t^x}{e^t} \leq \frac{-t^{\left\lceil x \right\rceil}}{e^t} \Rightarrow \lim \frac{-\left\lceil x \right\rceil !}{e^t} = 0$$

Would that be a valid argumentation? How else could I show that?

Here's another way: $$\lim_{t \to \infty} \frac{-t^x}{e^t} = -\left(\lim_{t \to \infty} \frac{t}{e^{t/x}} \right)^{x}$$ then apply L'Hôpital once.
$$-e^{-t}t^x = \frac {-t^x}{e^t} = -\frac {t^x}{\sum_{n=0}^{\infty} \frac {t^n}{n!}} =-\frac {t^x}{\sum_{n=0}^{[x]}\frac {t^n}{n!} +\sum_{n=[x]+1}^{\infty}\frac {t^n}{n!}}$$ where $[x]$ denotes least integer.
$$|\frac {t^x}{\sum_{n=0}^{[x]}\frac {t^n}{n!} +\sum_{n=[x]+1}^{\infty}\frac {t^n}{n!}}| < |\frac {t^x}{\sum_{n=[x]+1}^{\infty}\frac {t^n}{n!}}|<|\frac {t^x}{t^{[x]+1}}| \rightarrow 0$$
So, by comparison, $$|e^{-t}t^x|\to 0 \Rightarrow -e^{-t}t^x \to 0$$