If $P$ is an invertible linear operator on a finite dimensional vector space over a finite field, then there is some $n>0$ such that $P^n=I$ From Berkeley problems in Mathematics:

problem 7.4.21: let $P$ be a linear operator on a finite dimensional vector space over a finite field. show that if $P$ is invertible, then $P^n$ = $I$ for some positive integer $n$.

I think there is a mistake in the above statement: it does not hold true if P is a scalar multiple of the identity operator.
But is the statement correct otherwise? i.e. is the following modified statement correct?
let $P$ be a linear operator on a finite dimensional vector space over a finite field. assume $P$ is not a scalar multiple of the identity operator. show that if $P$ is invertible, then $P^n$ = $I$ for some positive integer $n$.

I'll make a try:
Take $P,P^2,P^3,...$. Then these can't be all different because the have entries from a finite field. Then $P^t=P^s\Rightarrow P^n=I$ since $P$ is invertible
