abstract algebra - What is wrong with this proof of Wedderburn's little theorem?

Wedderburn's little theorem $\quad$ every finite domain $A$ is a field.

Proof $\quad$ Let $x$ be a nonzero element of $A$. Because $A$ is finite, there
exist positive integers $n$, $k$ such that $x^n = x^{n + k}$. It is easy to
see by induction that the set $E = \left \{x^i : i \in \mathbf{N}^*\right\}$
does not contain $0$; it follows therefore from $x^n\left(1 - x^k\right) = 0$
that $x^k = 1$. Thus, $x^{k - 1}$ is the inverse of $x$ (when $k = 1$, $x$ has
inverse $1$).

All the proofs I have seen of this result are much more sophisticated than mine. Hence, I am doubting its correctness and could use a second opinion.

Answer

Wedderburn's little theorem states as you wrote above, that a finite domain is a field. A field is a commutative domain $A$, such that every nonzero $x \in A$ has a multiplicative inverse. Your proof only shows that any finite domain is a skew field. You must also prove that $A$ is commutative, which needs more sophisticated arguments.