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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

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    Perhaps try to extend your operations by considering both $q, m \in \mathbb{N}$ and then find values of $s^{p+q}$ and/or $s^{pq}$? – 2017-02-13
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    "I keep seeing these two descriptions being used interchangably when talking about idempotency". Would you have a precise reference where the second definition (for every $s\in S^1$ we can find $m\in\mathbb{N}$ s.t $s^m = 1$) is used? – 2017-02-13
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    Grillet's proof is perfectly correct, but your interpretation is wrong. You seem to assume that a monoid is always cancellative, which is not the case. In other words $x = ex$ does not imply $e = 1$. – 2017-02-13
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    @J.-E.Pin Sorry, meant to write definitions $s^2=s$ and $s^2=1$ being used. Clearly $s^2=1$ is meaningless in a semigroup, does that mean then we consider two different kinds of idempotencies? – 2017-02-13
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    They get some relation of the form $x = exz$, where $e$ is idempotent. Thus $ex = e(exz) = (ee)xz = exz = x$. – 2017-02-13
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    @J.-E.Pin Oh ok, now I get it :S Don't like laconic proofs that don't explain things properly :/ – 2017-02-13

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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.