How do I express 108 as the product of powers of its prime factors?

First notice that $108$ is even so $2$ is a factor: $$\begin{matrix} & & 108 \\ & \large\swarrow & & \large\searrow \\ 2 & & & &54\end{matrix}$$

$54$ is also even:

$$\begin{matrix} & & 54 \\ & \large\swarrow & & \large\searrow \\ 2 & & & &27\end{matrix}$$

$27$ is not even so let's check the next prime: $3$. We can see that $3$ divides $27$ because $2+7=9$ is a multiple of $3$. So:

$$\begin{matrix} & & 27 \\ & \large\swarrow & & \large\searrow \\ 3 & & & &9\end{matrix}$$

And then hopefully you can see that $9$'s prime factorization is just $3\times 3$. So putting all of these steps together we see that

$$\begin{matrix} & & 108 \\ & \large\swarrow & & \large\searrow \\ \color{red}2 & & & & 54 \\ & & & \large\swarrow & & \large\searrow \\ & & \color{red}2 & & & & 27 \\ & & & & & \large\swarrow & & \large\searrow \\ & & & & \color{red}3 & & & & 9 \\ & & & & & & & \large\swarrow & & \large\searrow \\ & & & & & & \color{red}3 & & & & \color{red}3\end{matrix}$$

Thus

$$108 = 2\times 2 \times 3 \times 3\times 3 = 2^2\times 3^3$$

Yes you are correct.

Just divide 108 by its prime factors and count them !

$\frac{108}{2}=54, \frac{54}{2}=27 \Rightarrow$ $2^2$

$\frac{27}{3}=9, \frac{9}{3}=3 ,\frac{3}{3}=1 \Rightarrow 3^3$

So $108=2^2\cdot3^3$