Answer: 55

If $n$ is even, then $n$ can be written as $2m$ for some integer $m$.

Then, using indicies laws: $n^n = n ^{2m} = \left(n^m\right)^2$, which is a square number.

If $n$ is odd, then $n$ can be written as $2r + 1$, for some integer $r$.

Then, $n^n = n^{2r + 1} = n \times n^{2r} = n \times \left(n^r\right)^2$.

For $n^n$ to be a square, $n$ must also be a square.

If $n$ is indeed a square number, $n = s^2$ for some integer $s$.

Then, using indices laws: $n^n = \left(s^2\right)^n = s^{2n} = \left(s^n\right)^2$, which is a square number.

Hence we need to count all the even numbers and all the odd squares. There are $50$ even numbers between $1$ and $100$, and $5$ odd squares.

This means there are $55$ numbers $n$ such that $n^n$ is a square.