Answer: 3

Using powers of $2$

$4 = 2^2$ and $8 = 2^3$. Therefore:

$4^4 = \left(2^2\right)^4 = 2^{2 \times 4} = 2^8$

$8^8 = \left(2^3\right)^8 = 2^{3 \times 8} = 2^{24}$

This means: $8^8 = 2^{24} = 2^{8 \times 3} = \left(2^8\right)^3 = \left(4^4\right)^3$.

Hence, $4^4$ needs to be cubed to obtain $8^8$.

Using powers of $4$

$\begin{align}8^8 &= \left(2\times4\right)^8\\

&=2^8\times4^8\\

&=2^{2\times4}\times4^8\\

&=\left(2^4\right)^4\times4^8\\

&=4^4\times4^8\\

&=4^4\times4^4\times4^4\\

&=\left(4^4\right)^3\end{align}$