Let $S$ be a (finite) semigroup.
An element $s\in S$ is idempotent iff $s^2=s$
One can also show, if $S$ is finite, then for every $s\in S$ there exists $n\in\mathbb{N}$ s.t $s^n$ is idempotent i.e $s^n = s^{2n}$.
When we operate in $S^1$ (a monoid), does idempotency (identity potency) mean that for every $s\in S^1$ we can find $m\in\mathbb{N}$ s.t $s^m = 1?$.
E: confusion resolved
No, an idempotent in a monoid does not have to be 1. Trivial example: Consider {0, 1} under standard multiplication, which is clearly a monoid - we have both 0*0=0 and 1*1=1, so both 0 and 1 are idempotent, but only 1 is an identity.