Does the series $\sum_{n=0}^{\infty} \frac{(n^n)!}{n^{n!}}$ converge or diverge? I don't know if the series
\begin{equation}
\sum_{n=0}^{\infty} \frac{(n^n)!}{n^{n!}}
\end{equation}
converges or diverges, and it fits no simple test, so I am completely stuck. Is there some unknown test that can be used to solve this?
Edit: Does 
\begin{equation}
\sum_{n=0}^{\infty} \frac{n^{n!}}{(n^n)!}
\end{equation} 
converge and how? 
(Sorry if this is against the rules, but I want to see how to use another test for a series like this._
 A: Note $$(n^n)!\geq n(n+1)\dots (n^n-1)(n^n)\geq (n)(n)\dots (n)(n)=n^{n^n-n+1}$$
You can prove that $ n^n-n\geq n!$ (for $n$ big enough).
So, $(n^n)!\geq n^{n^n-n+1} \geq n^{n!+1}$. This proves that $(n^n)!$ grows faster than $n^{n!}$. 
So, by divergence test, your series diverges.
A: For this kind of problem it's useful to compare the logarithms of the numerator and denominator.  The log of the denominator is $n!\log{n}$ but for the numerator we need to estimate $\log{n!}$. Now $$\log{n!}=\log{1}+\log{2}+\cdots+\log{n}=\sum_{k=2}^n\log{k}$$  Since $\log$ is increasing we have $$\int_{k-1}^k\log{x}\mathrm{dx}<\log{k}<\int_k^{k+1}\log{x}\mathrm{dx}$$ so that $$\int_1^n\log{x}\mathrm{dx}<\log{n!}<\int_2^{n+1}\log{x}\mathrm{dx}$$  Doing the integration we end up with $$n\log{n}-n<\log{n!}<(n+1)\log{n+1}$$
Applying this to our problem, we se that the logarithm of the numerator is greater than  $$n^n\log(n^n)-n^n=n^{n+1}\log{n}-n^n>n^{n+1}(\log{n}-1)$$  Now  the ratio of this last expression to the value of the logarithm of the denominator, is
$${n^{n+1}(\log{n}-1)\over n!\log{n}}$$ and obviously this ratio goes to $\infty$ as $n\to\infty.$ 
Not only does the original fraction go to $\infty$, but the ratio of the logarithm of the numerator to the logarithm of the denominator goes to $\infty.$  We might say that the original expression goes to $\infty$ in an indecent hurry! 
A: Even the limit is not zero (infinity, in fact) at infinity, so it diverges...
