The title means that if you raise a number to a certain power, and then use raise the resulting number to another power, what do you get?
In symbols, we get \[\left(a^m\right)^n = a^{mn}\,.\]
The \(a^m\) expression inside means this: \[\underbrace{a \times a \times \dotsb \times a}_{\text{\(a\) appearing \(m\) times}}\,.\]
The \(({\phantom{a}})^n\) part means to repeat “\(a^m\)” \(n\) times: \[ \text{repeat \(n\) times} \left\{ \vphantom{ \begin{align} &a\\ &a\\ &\vdots\\ &a \end{align} } \right. \begin{align} &\overbrace{a \times a \times \dotsb \times a}^{\text{\(m\) times}}\\ \times &a \times a \times \dotsb \times a\\ &\phantom{a \times a}\vdots\\ \times &a \times a \times \dotsb \times a\\ \vphantom{a} \end{align} %\right. \]
So \(\left(a^m\right)^n\) is the same as multiplying \(n\) rows of \(m\) lots of \(a\), with the total of \(m \times n\) lots of \(a\) being multiplied together.
This is the same as raising \(a\) to the power of \(m+n\).
Take \(13^4\) as an example. This is the same as \[13 \times 13 \times 13 \times 13.\]
Now, raise that whole thing to the power of \(2\) and it’ll become \begin{align} &(13 \times 13 \times 13 \times 13)\\ \times &(13 \times 13 \times 13 \times 13) = 13^{4 \times 2} \end{align}
If instead of squaring, you cubed it: \begin{align} &13 \times 13 \times 13 \times 13\\ \times &13 \times 13 \times 13 \times 13\\ \times &13 \times 13 \times 13 \times 13 = 13^{4 \times 3} = 13^{12} \end{align}
And, to the power of five: \begin{align} &13 \times 13 \times 13 \times 13\\ \times &13 \times 13 \times 13 \times 13\\ \times &13 \times 13 \times 13 \times 13\\ \times &13 \times 13 \times 13 \times 13\\ \times &13 \times 13 \times 13 \times 13 = 13^{4 \times 5} = 13^{20} \end{align}