Find the minimum value $f(x) = \dfrac{e^x}{x^n}$ for $x>0$ 
Find the minimum value $f(x) = \dfrac{e^x}{x^n}$ for $x>0$ and conclude that $f(x) > \dfrac{e^n}{n^n}$ for $x > n$.

I found the derivative $\dfrac{d}{dx}\left(\dfrac{e^x}{x^n}\right) = e^x x^{-n-1} (x-n)$ and set it equal to zero to find that $x = n$ is the solution. Then I found the second derivative, which is $e^x x^{-2-n} (n+n^2-2 n x+x^2)$, and plugged in $x = n$ to find that it was positive implying that this is a global minimum.
 A: A preliminary remark: note that the result is false for $n<0$ (because $f(x)$ assumes arbitrarily large negative values for small negative $x$ if $n$ is odd). You might think $n>0$ was implicit, but a careful reading shows it is not. [Thanks to @BarryCipra] 
However, let us assume that $n>0$.
Yes, your proof is partially correct. To finish off you need to note that since $f(x)$ has a global minimum at $x=n$ we have $f(x)>f(n)$ for all $x>0$. [Thanks to @user1892304]
But it is not enough to find that the 2nd derivative is positive at a zero of the first derivative. That might or might not indicate a global minimum. To establish a global minimum at $x=n$ you need that the 1st derivative is $<0$ for $x<n$ and $>0$ for $x>n$. It is easy to see directly from the expression for the 1st derivative that this is true here. [Thanks to @AndreNicholas]
Alternatively, note that $n+n^2-2nx+x^2=n+(n-x)^2>0$ so that $f''(x)>0$ for all $x$, not just $f''(n)>0$. [Thanks to @BarryCipra]
The plot shows the case $n=3$
 
