Prove that $f'(x) > \dfrac{e^{n+1}}{(n+1)^{n+1}}$ for $x > n+1$ 
Let $f(x) = \dfrac{e^x}{x^n}$. Using $f'(x) = \dfrac{e^x(x-n)}{x^{n+1}}$, prove that $f'(x) > \dfrac{e^{n+1}}{(n+1)^{n+1}}$ for $x > n+1$, and thus obtain another proof that $\displaystyle \lim_{x \to \infty} f(x) = \infty$.

I was thinking of taking the second derivative but that would get computational. Is there another way to solve this?
 A: The second derivative is not that computational. 
One may set, for $x>0$, $n>0$:
$$
g(x)=f'(x)-\dfrac{e^{n+1}}{(n+1)^{n+1}}=\dfrac{e^x(x-n)}{x^{n+1}}-\dfrac{e^{n+1}}{(n+1)^{n+1}}
$$ giving, with the use of $(uv)'=u'v+uv'$:
$$
\begin{align}
g'(x)&=-(n+1)\frac1{x^{n+2}}\cdot e^x(x-n)+\frac1{x^{n+1}}\cdot e^x(x-n+1)
\\\\&=\frac{e^x}{x^{n+2}}\cdot((x-n)^2+1)
\\\\&>0.
\end{align}
$$ Thus $g$ is strictly increasing over $(0,\infty)$: $g(x)>g(n+1)$ for all $x>n+1$, but $g(n+1)=0$ then 

$$
\dfrac{e^x(x-n)}{x^{n+1}}-\dfrac{e^{n+1}}{(n+1)^{n+1}}>0 \quad \text{for} \quad x>n+1.
$$

A: We have $f'(x)=(1-\frac{n}{x})f(x)$. Also $f'(n+1)=\frac{e^{n+1}}{(n+1)^{n+1}}$. Clearly $f(x)>0$ for positive $x$ and $1-\frac{n}{x}$ is positive for $x>n+1$, so $f'(x)$ is positive for $x>n+1$. Hence $f(x)$ is increasing for $x>n+1$ and so is $(1-\frac{n}{x})$. Hence $f'(x)$ is increasing for $x>n+1$ and so $f'(x)>f(n+1)$ for $x>n+1$. 
A: Once you find the 2nd derivative you can show its always positive and thus the "edge" x=n+1 is an absolute minimum, hence f'(x)>f'(n+1)
A: Since $(x-n)$ is strictly increasing and positive, all we have to do is show $e^x/x^{n+1} = e^xx^{-(n+1)}$ is strictly increasing on $[n+1,\infty).$ The derivative of this is
$$e^x x^{-(n+2)}(x-(n+1)),$$
which is positive on $(n+1,\infty)$ and we're done.
