Why does $(\log n)^{\log n} = \Omega(n^{10})$?

Question

Why does $(\log n)^{\log n} = \Omega(n^{10})$?

In other words, show that $(\log n)^{\log n} \ge c\cdot n^{10})$ for some constant $c>0$.

I'm not sure how to prove it, how can I write $(\log n)^{\log n}$ in a simpler way?

Answer 1

Hint: take logarithm of both sides

Answer 2

Note that $$(\log n)^{\log n}=\exp(\log n\cdot \log(\log(n)))=n^{\log(\log(n))}$$ Hence $(\log n)^{\log n} = \Omega(n^{a})$ for any $a>0$ because $\log(\log(n))$ goes to infinity and it is therefore bigger than any constant.