finite dimensional division algebra $D$ we know that a nonzero unital ring  $D$ in which every nonzero element is
invertible (i.e., $D^{*} =  D - \{ 0 \} $) is called a division ring.Division rings obviously
have no (left, right) zero-divisors. A commutative division ring is called a field.
But I got into trouble and confused how to prove the following. Is the question at all correct? Help me please.
my problem is:

A finite dimensional division algebra  $D$  is a finite field if and only
  if there exists $ n \in \mathbb{N}$  such that $x^{n} = x $ for all $ x \in D $

 A: This is more or less a previous step in the proof of the famous theorem of Jacobson that states that a ring which satisfies $x^{n_x}=x$ for every element $x$ is commutative (see Structure theory for algebraic algebras of bounded degree (Jacobson, 1945; Theorem 11)).
First of all, $D$ has prime characteristic, since $2^n=2$ (this can not happen in characteristic $0$). Suppose $D$ is not commutative; then there exists $a\in D\setminus Z(D)$. The fundamental step is that there exists $b\in D$ such that $bab^{-1}=a^i$, with $a^i\neq a$ (see for example Lemma 3.1.1. of Noncommutative rings (Herstein, 1971 (2ed)). Then, if $P$ is the prime field of $D$, $P(a,b)$ is a finite ring because $a^n=a, b^n=b, a^ib=ba$, and it is a division ring because $x^{-1}=x^{n-2}\in P(x)$ for every $x\in D$. Therefore $P(a,b)$ is a finite division ring, that should be commutative by Wedderburn's little theorem, but isn't, as $ba=a^ib\neq ab$. This shows by contradiction that there is no such $a$, i.e., $D=Z(D)$ is commutative.
Now every element of $D$, which is a field, is a root of the polynomial $X^n-X$, so $D$ has at most $n$ elements.
A: Ok, this way is better for our purposes (using finite dimension):
Let $D$ be a finite-dimensional division algebra over the field $K$ such that $x^n=x$ for every $x\in D$. Consider the center $Z(D)$ of $D$, which is a field and contains a copy of $K$. Since every $x\in Z(D)$ is a root of the polynomial $X^n-X$, $Z(D)$ is a finite field, so $K\subseteq Z(D)$ is a finite field. Therefore, since $D$ is finite-dimensional over $K$, $D$ is also finite (as there is only a finite number of linear combinations of basis elements). Now apply Wedderburn's little theorem: every finite division ring is a field.
For the converse statement, pick a finite field $F$ of order $q=p^m$, $p$ a prime. Then the multiplicative group $F^*$ has $q-1$ elements. By Lagrange's theorem, the order of every element of the group divides $q-1$. Therefore $x^{q-1}=1$ for every $x\in F^*$ and hence $x^q=x$ for $x\in F^*$. But this identity is also true for $x=0$, so it is true for every $x\in F$.
