Upper bound for $e^{-x}$

Question

Trying to show for all $n$ there is a positive $x_0>0$ such that for all $x\geq x_0$: $$e^{-x}\leq \frac{n!}{x^n}$$

The function $e^{-x}-\frac{n!}{x^n}$ seems to be negative for all $x>0$ for all the values of $n$ I've tried but I'm sure there's a more straight forward argument using Taylor expansion that I'm not seeing.

Answer 1

From the series expansion $e^{x} \geq \frac {x^{n}} {n!}$ for all $x >0$. This inequality is the same as the one you are trying to prove. Any $x_0>0$ is fine.

Answer 2

Over $\mathbb{R}^+$, the only stationary point of $f(x)=x^n e^{-x}$ occurs at $x=n$, since this is the only solution of $f'(x)=0$. This stationary point is an absolute maximum, hence for any $x>0$ we have $$ x^n e^{-x} \leq \frac{n^n}{e^n}. $$ This is an improvement of $x^n e^{-x}\leq n!$, since $\frac{n^n}{e^n}<n!$ can be proved via the Hermite-Hadamard inequality (among other ways). We have $$ \frac{1}{2}\log(1)+\log(2)+\ldots+\log(n-1)+\frac{1}{2}\log(n) < \int_{1}^{n}\log(x)\,dx = n\log n-n+1 $$ and $$ \log(2)+\log(3)+\ldots+\log(n-1) > \int_{3/2}^{n-1/2}\log(x)\,dx $$ so $$ \log(n!) > \log(n) + \left(n-\tfrac{1}{2}\right)\log\left(n-\tfrac{1}{2}\right) -n+2-\tfrac{3}{2}\log\tfrac{3}{2}>\left(n+\tfrac{1}{2}\right)\log n-n+1$$ which gives $n! > \frac{n^n}{e^n}\sqrt{en}$.