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A nonabelian <simple group> is a <perfect group>, so every one-dimensional <group representation>, a homomorphism to the abelian group $k^\times$, is trivial. Thus a nontrivial irreducible <Brauer character> cannot have degree $1$.
Suppose instead that its representation has dimension $2$, working over a splitting extension if necessary. Its determinant is a one-dimensional representation and is therefore trivial. Its kernel is a <normal subgroup>, and the representation is nontrivial, so simplicity makes it faithful. Consequently $G$ embeds in $\operatorname{SL}_2(k)$.
We use the <Feit–Thompson theorem>: every finite group of odd order is solvable. Therefore this nonabelian simple group has even order, and <Cauchy's theorem for finite groups> supplies an <involution> $t$. In odd <characteristic>, its representing matrix $A$ satisfies $A^2=I$ and is diagonalizable with <eigenvalues> in $\{1,-1\}$. Since $\det A=1$, it is $I$ or $-I$. Faithfulness excludes $I$, so $t$ acts as the scalar matrix $-I$. It commutes with the whole image; faithfulness then makes $t$ central in $G$, contradicting nonabelian simplicity. Hence $\boxed{\chi(1)>2}$.
The <odd order theorem> is the deep standard group-theoretic input in this proof; its use is explicit rather than hidden in an unsupported assertion that $G$ has an involution.
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