Show that $e^{-x}x^n$ is bounded on $[0,\infty)$ and hence prove that $\int_0^\infty e^{-x}x^n \, dx$ exists. Show that $e^{-x}x^n$ is bounded on $[0,\infty)$ for all positive integral values of $n$. Using this result show that $\int_0^\infty e^{-x}x^n \, dx$ exists.
My work:
I know $f$ is a continuous function and $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{x^n}{e^x}=\dfrac{n!}{e^x}=0$, by applying L'hospital's rule repeatedly.
But how to prove its boundedness formally? Also for existence of an integration, my book only covers for those integration where the limits are finite (Darboux's condition for integrability). So, how do I prove existence of an integration if the limits are infinite.

$f(x)=e^{x-1}$ is convex ($f''(x)>0$ for all real $x$), so its graph lies above the tangent at $x=1$: $e^{x-1}\ge x$ for all real $x$. This means $x\,e^{-x}\le e^{-1}$. Replacing $x$ by $x/n$ and raising to the $n$th power gives $$x^n\,e^{-x}\le n^n\,e^{-n}=C_n$$ for $x>0,$ that's the boundedness.
We have to show only the existence of the improper integral $\int^\infty_1x^n\,e^{-x}\,dx,$ since the existence of $\int^1_0x^n\,e^{-x}\,dx$ is trivial.
But $$\int^b_1x^n\,e^{-x}\,dx=\int^b_1\frac{x^{n+2}\,e^{-x}}{x^2}\,dx\le \int^b_1\frac{C_{n+2}}{x^2}\,dx\le\int^\infty_1\frac{C_{n+2}}{x^2}\,dx=C_{n+2},$$  because $x^{n+2}\,e^{-x}\le C_{n+2},$ as was shown above. The LHS is monotone (the integrand is positive) and bounded, so the limit as $b\rightarrow\infty$ exists.
