TITLE: Dummit and Foote, question about ex. 7.28(d). Is my alternative proof correct?
QUESTION [3 upvotes]: Let $R$ be a ring with $1 \neq 0$. A nonzero element $a \in R$ is called a left zero divisor in $R$ if there is a nonzero element $x \in R$ such that $ax = 0$. Symmetrically, $b \neq 0$ is called a right zero divisor in $R$ if there is a nonzero element $y \in R$ such that $yb = 0$. (So a zero divisor is an element which is either a left or a right zero divisor, or both.) An element $u \in R$ is said to have a left inverse in $R$ if there is some $s \in R$ such that $su = 1$. Symmetrically, $v$ has a right inverse if there exists $t \in R$ such that $vt = 1$.
(d) Prove that if $R$ is a finite ring then every element that has a right inverse is a unit (i.e. has a two sided inverse).

Solution can be found here https://web.archive.org/web/20150512111012/https://crazyproject.wordpress.com/2010/08/14/basic-properties-of-left-and-right-units-and-left-and-right-zero-divisors/
I solved this question differently and want to know if my solution is correct?
My solution: Let $n$ be number of elements in $R$. Let $u\in R$ have a right inverse $r\in R$: $ur=1$. Consider the set $S=\{u,u^2,u^3,..., u^{n+1}\}$. Since $R$ is a ring, we have $S\subseteq R$. According to pigeonhole principle, there are at least two elements in $S$ equal to each other: $u^k=u^l$, $1\le l<k\le n+1$. Multiplying both sides by $r$ from the right $l$ times we get
$$
u^m=1, \,\,\,\, 1\le m=k-l\le n.
$$
$u$ commutes with itself, thus we have
$$
u\cdot u^{j}=u^{j}\cdot u=1,\,\,\,\, j\ge 0.
$$
This proves that $u$ is a unit. $\Box$
Is this proof correct and acceptable?

REPLY [2 votes]: Yes, that is a perfectly valid proof. 
In fact, the two proofs are closely related (the pigeonhole principle is closely related to the fact that an injective endomap of a finite set is surjective)
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