The maximum of xⁿe^(−x)

1. $f'$ is positive on $(0,n)$ and negative on $(n,\infty)$, so $f$ attains its maximum at $x=n$:
$$\max f=f(n)=n^ne^{-n}=\left(\frac ne\right)^n.$$

2. $x^ne^{-x}\le\left(\frac ne\right)^n$, i.e. $e^x\ge\left(\frac en\right)^nx^n$.

3. For $n=2$: $e^x\ge\frac{e^2}4x^2\approx1.85\,x^2$, with equality only at $x=2$. Compare with the series bound $e^x\ge1+x+\frac{x^2}2$: for large $x$ the new bound is better (it grows like $1.85\,x^2$ against $0.5\,x^2$), while for small $x$ the series bound is better (the new one tends to $0$).

Remark. Part 2 shows that $e^x$ dominates every power of $x$ in a quantitative way: $\dfrac{e^x}{x^n}\ge\left(\dfrac en\right)^n>0$ for all $x>0$, and the bound is sharp.