there is a binomial formula:
$(x+y)^n=\displaystyle\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$
When operations are done in $GF(2^m)$ then all positive integers are reduced $\bmod2$, so binomial formula for $n=2^i$ in $GF(2^m)$ is:
$(x+y)^{2^i}=x^{2^i} + y^{2^i} $
So now the question. If there are reduced all binomial coefficients $\binom{n}{k}$, then why exponents like $2^i$ are not reduced?