TITLE: $R$ is a finite ring and for every $a \in R$, there exists a natural number $n(a)$ such that $a^{n(a)}=a$
QUESTION [6 upvotes]: $R$ is a finite ring and for every $a\in\,R$ there exist natural number  $n(a)>1$ that  $a^{n(a)}=a$ . Is $R$ a ring with identity? 

If this question is correct then, for every $a \in R\,\,,a^{n(a)-1}$ is identity of $R$. I prove that $a^{n(a)-1}\in Z(R)$. But how I can prove that for every $a \in R\,\,,a^{n(a)-1}$ is identity of $R$?  
I find a proof for my quastion as follows :
Let $R=\{a_1,a_2,...,a_n\}$.
We know for any $a_i$ there exit $n_i$ s.t $a_i^{n_i}=a_i$ and $a_i^{n_i-1}\in Z(R)$.
It is enough to show that there exit a some element $e$ s.t $ea_i=a_ie=a_i$.
I'm going to construct an identity's element.
Consider:
 $e_1=a_1^{n_1-1}$ $\Rightarrow \, e_1a_1=a_1e_1=a_1$
$e_2=e_1+a_2^{n_2-1}-e_1a_2^{n_2-1}\Rightarrow \, e_2a_1=a_1e_2=a_1 \,\,, \,\,e_2a_2=a_2e_2=a_2$
$e_3=e_2+a_3^{n_3-1}-e_2a_3^{n_3-1}\Rightarrow \, e_3a_1=a_1e_3=a_1 \,\,, \,\,e_3a_2=a_2e_3=a_2\,\,, \,\,e_3a_3=a_3e_3=a_3$
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$e_n=e_{n-1}+a_n^{n_n-1}-e_{n-1}a_n^{n_n-1}\Rightarrow$ for every $0\leq i\leq n$ we have $e_na_i=a_ie_n=a_i \Rightarrow \, 1_R=e_n$.

REPLY [3 votes]: I think the famous theorem of Jacobson about such a ring being commutative holds, even if there is no identity. 
The hypothesis implies the ring is von Neumann regular. Therefore every ideal is idempotent, including the entire ring. 
Since the ring is finite, by Nakayama's lemma, we have that R has an identity.
It would follow that $R$ is a finite product of fields.