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.

For the boundedness, yoou can take the derivative and show that it is negative on some interval $[a,\infty)$ and thus show the function is bounded on $[a,\infty)$. It is bounded on $[0,a]$ since it's continuous.
The way you show the integral on the infinite interval exists is to remember the definition of an improper integral existing: $$ \lim_{b\to\infty}\int_0^b x^n e^{-x} dx$$ exists and is finite. It should be said that bounded on $[0,\infty)$ does not imply that the integral exists (take $\frac{1}{x+1}$ for instance), so that part of the question seems a bit misleading. However, it's true here and a simple way to prove it is by induction. You can show the integral exists for $n=0$ straightforwardly, and then for the induction step, use integration by parts.
