induction for idempotent matrix : $P^n = P$ 
Given that $P^2 = P$ how do i prove by induction that $P^n = P$?

I have tried the following:  we know that $P^k = P$ holds for $k = \{1,2\}$. If we now take $k=3$:
$$
\begin{align}
P^3 &= P^2P
\\ &=PP \tag*{($P$ is idempotent)} \\
\\&= P^2
\\&=P
\end{align}
$$
therefore $P^k = P$ holds for all natural numbers.
however, this seems... incomplete for me... Am I missing something? 

Suppose $P^{n-1}=P$.
Then
$$\begin{align*}P^n&=P(P^{n-1})\\
&=PP\\
&=P^2\\
&=P.
\end{align*}
$$
We're given $P^2=P$, so by induction on $n$, we're done.
Thinking of induction as reaching back to the previous cases instead of reaching forward to the next case can be insightful.
