Let \[f(x) = \left\{
\begin{array}{cl}
\sqrt{x} &\text{ if }x>4, \\
x^2 &\text{ if }x \le 4.
\end{array}
\right.\]Find $f(f(f(2)))$.
To evaluate $f(f(f(2)))$, I need to start from the inside and work my way out. First, I need to find $f(2)$. Since $2 \le 4$, I use the second rule of $f$, which says $f(x) = x^2$. So, $f(2) = 2^2 = 4$. Next, I need to find $f(f(2))$, which is $f(4)$. Since $4 \le 4$, I still use the second rule of $f$, which says $f(x) = x^2$. So, $f(4) = 4^2 = 16$. Finally, I need to find $f(f(f(2)))$, which is $f(16)$. Since $16 > 4$, I use the first rule of $f$, which says $f(x) = \sqrt{x}$. So, $f(16) = \sqrt{16} = 4$. Therefore, $f(f(f(2))) = 4$. 