Tricky logarithm algebra question

I need to solve for h in the following equation:
log_2(n^(1/(2^h))=1 (n and h are both positive by the way)

I made both sides an exponent of 2 which got me to:
n^(1/(2^h))=2

I need to keep hacking away at that left side to get h by itself but I'm not sure how to proceed. Can I take log_n of both sides?

n^(1/2^h) = 2
(1/2^h) * ln(n) = ln(2)
2^h / ln(n) = 1/ln(2)
2^h = ln(n) / ln(2)
2^h = log[2](n)
h * log(2) = log(log[2](n))
h = log(log[2](n)) / log(2)
h = log[2](log[2](n))