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I was just looking at the following limit

$$ \lim_{n \to \infty} \frac{2^{2^{n}}}{n^{2n}}. $$

I can guess that the limit is $\infty$, and plotting a graph of the numerator and denominator confirms my suspicion. I'm struggling for a straightforward way to show this mathematically, however, which would be rather nice. Any ideas?

I've attempted to use $\log$ however to do so I have to assume the limit exists, which in this case doesn't work. I also tried finding a suitable inequality without much luck.

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    Yes. Roughly, that's because $n^n$ is $o(\text{double exponential})$. – 2017-01-23

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No derivatives required here: just show the log tends to $\infty$. Indeed $$\ln\frac{2^{2^n}}{n^{2n}}=2^n\ln 2-2n\ln n,$$ and $n\ln n=o(n^2)$, $\;n^2=o(2^n)$, so $n\ln n=o(2^n)$.