Prove that every finite domain contains an identity element 
Prove that every finite domain contains an identity element.

Please give me help 
 A: Let $a$ be a non-zero element of the domain. Consider the objects $a$, $a^2$, $a^3$, and so on.
The powers of $a$ cannot all be different, since if they were, there would be infinitely many elements in the domain. It follows that there are natural numbers $m$ and $n$, with $m<n$, such that $a^m=a^n$.  Thus for any $x$ in the domain,
$$xa^n=xa^m,$$
and therefore
$$xa^{n-m}a^m=xa^m. \qquad(\ast)$$
Since $a\ne 0$, we have $a^m\ne 0$, so by $(\ast)$ and the cancellation property,
$$xa^{n-m}=x.$$
This says that $a^{n-m}$ is a multiplicative identity.
A: Hint $\:$ The elements $\ne 0$ form a nonempty finite semigroup, so contain an idempotent $\rm\:e.\:$ Thus
$$\rm\: 0\: \ne\: e\: =\: e^2\: \ \Rightarrow\ \ e\:x\: =\: e^2\:x\ \ \Rightarrow\ \ x\: =\: e\:x$$
Alternatively, notice that $\rm\  a\ne 0\:\Rightarrow\: x\mapsto a\:x\ $ is $1$-$1$ so onto, therefore 
$\qquad\qquad$ for all $\rm\:x\!:\ $
$\rm\begin{eqnarray}\exists\: e\!:\ \ a\: &=&\:\rm \color{#C00}{a\:e}\\ \rm \exists\: d\!:\ \ x\: &=&\:\rm a\:d\end{eqnarray}$ $\ \ \Rightarrow \ \ \begin{eqnarray}\rm a\:d \: &=&\: \rm \color{#C00}{e\:a}\:d\\ \rm x \:&=&\: \rm e\:x\end{eqnarray} $
The proof in André's answer is a special case of the first method above. Probably the proof of Herstein mentioned by Chandrasekhar is similar to the second proof above.
A: This is a property that every finite integral domain $D$ is a field. Please look into I.N.Herstein's text.

Idea. Is to consider a non-zero element in $D$ and establish a bijection from $D \to D$ via $x \mapsto ax$.
Please look into Page 128 of the following link
