TITLE: What would be complexity of computing $3^{n^n}$?
QUESTION [5 upvotes]: Just curious, what would be the computational complexity of computing $3^{n^n}$?
I am not sure what it would be like.

REPLY [2 votes]: Let $M(n)$ denote the complexity of the multiplication of two integer with $n$-digits (let say in base two).
Thanks to Schonhage and Strassen we know that $M(n) = O(n \log n \log\log n)$. Thanks to Fürer, we even know that it's a bit lower.
Let $a$ be an integer and $b$ an integer. Then $a^b$ can be computed in time $O(M(b \log_2 a))$ using fast exponentiation. This is not bad: since $a^b$ has bit-size exactly $b\log_2 a$, the cost of the computation is almost the size of the output.
So the complexity of computing $a^{b^c}$ is the cost of computing of $b^c$ plus the cost of computing $a^{b^c}$. When $a$ is at least 2, then the second operation dominate the first, and the total cost is $O(M(b^c \log_2 a))$.
We could make explicit the constant in the big-oh, and we would realize that is is not much larger than 2 or 3.