How is $ss^{-1}$ idempotent in an inverse monoid?

Question

An inverse monoid S is a monoid such that for all $s \in S$, there exists a $t \in S$ such that $s=sts$ and $t=tst$. In this case, we write $t = s^{-1}$.

Why is $ss^{-1}$ an idempotent? I don't understand how $(ss^{-1})^2 = ss^{-1}ss^{-1} = ss^{-1}$ when $ss^{-1}$ is not guaranteed to be the identity element.

Answer 1

$(ss^{-1})^2=ss^{-1}ss^{-1}=(ss^{-1}s)s^{-1}=ss^{-1}$, where the last step is according to the definition of inverse.

Answer 2

I'll keep the notation $t$ for $s^{-1}$. You have to prove $(st)(st)=st$. But $$(st)(st)=(sts)t=st. $$