Why is RMS used for P=VI (where there are no squares)?
In short: Using RMS values for AC power yields a simpler expression that is analogous to the expression for DC power of \$P = IV\$.

The expression for AC power, using peak values for current and voltage, is:
$$
P = \frac{1}{2} \cdot I_{pk} \cdot V_{pk}
$$

RMS values simplify this by absorbing the \$\frac{1}{2}\$ factor into the current and voltage values:

$$
\begin{align}
P &amp;= \frac{1}{2} \cdot I_{pk} \cdot V_{pk}\\
P &amp;= \frac{1}{\sqrt{2}}I_{pk} \cdot \frac{1}{\sqrt{2}}V_{pk}\\
P &amp;= (\frac{I_{pk}}{\sqrt{2}}) \cdot (\frac{I_{pk}}{\sqrt{2}})\\
P &amp;= (I_{RMS}) \cdot (V_{RMS})\\
\end{align}
$$

Note that the equations above are only representing real power across a purely resistive load. Of course, with a reactive load, you'll need to work with complex power, but from your question it seems you haven't started studying that just yet.