TITLE: Short proof that $X^2 = X \Rightarrow X^{100} = X$
QUESTION [5 upvotes]: Given that a matrix $X$ satisfies $X^2 = X$ it is clear that $X^{100}=X$ by repeated multiplication of $X$. Algebraically, we might write:
$$X^{100} = (X^2)^{50}=X^{50}=(X^2)^{25}=X^{25}=X(X^2)^{12} = \dots = (X^2)^2 = X $$
But this seems like too much work for such a simple fact. Is there a short algebraic proof?

REPLY [3 votes]: Well, since $X^2 = X$, i.e., $X$ is idempotent, the standard inductive proof for idempotents in any ring works:  the inductive hypothesis is $X^k = X$, whence $X^{k + 1} = XX^k = XX = X^2 = X$.  Take $X^2 = X$ the base case and we're done, we have $X^n = X$ whether $n = 10$, 
$n = 10^6$, or $n$ is one-gazillion or even more!  What fun!  Cheers!