Prove if an element of a monoid has an inverse, that inverse is unique

I'm in the beginning stages of this proof. My question here I guess is that by definition a monoid has the properties that it is associative and has an identity. So: $(ab)c=a(bc)$ and $de=ed=e$ where $e$ is the identity. If I can prove that the identity is unique, does that prove the inverse is unique?

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No, proving the identity is unique would only prove just that. To prove that inverses are unique when they exist, you can suppose that some $a$ in your monoid has inverses, say $b$ and $c$, and show that $b=c$. – Jonas Meyer Jan 27 '12 at 5:01

Of course the identity is unique, $e_1=e_1\circ e_2=e_2$. Now suppose $a,b$ are both inverses of $x$. Then
$$a = a\circ e = a\circ(x\circ b) = (a\circ x)\circ b = e\circ b = b.$$
@Myself: How is it necessary? Just let $e$ be any identity (not that there are any more than one but let's pretend we don't know this) and the line given here is still valid - without assuming or otherwise showing uniqueness of the identity. – anon Jan 27 '12 at 19:56