Showing that $\log(\log(n))^{\log(n)}$ is $O(7^{\sqrt n})$

Question

What's a straightforward way to prove that $\log(\log(n))^{\log(n)}$ is $O(7^{\sqrt n})$?

(I'm dealing with Big O Notation)

Answer 1

Start with

$$7=(\log(\log(n)))^{\log_{\log(\log(n))}(7)} = (\log(\log(n)))^{\log(7)/\log(\log(\log(n)))}.$$

So

$$7^{\sqrt{n}} = (\log(\log(n)))^{\log(7) \sqrt{n}/\log(\log(\log(n)))}.$$

Now you just have to prove that the exponent is asymptotically larger than $\log(n)$.

This is a good way to start a lot of these sorts of problems: the idea was to make one expression look more like the other expression, so as to reduce the size of the problem.

Answer 2

Just use the fact that $\ln(n) < n$, so $\ln(\ln(n)) < n$ (for n big enough)

Then

$$\ln(\ln(n))^{\ln(n)} = e^{\ln(n) \ln( \ln (\ln(n))) } \leq e^{ \ln(n) \ln(n) } $$

And as $\sqrt{n} \geq (\ln(n))^2$ (i assume this is known),

$$\ln(\ln(n))^{\ln(n)}\leq e^{\sqrt{n} \ln(7) } = 7^{\sqrt{n}}$$