Let $E/F$ be a finite extension of finite fields; hence, it is a cyclic Galois extension, so let the Galois group be $G$. Hilbert's Theorem 90 states that $H^1(G, E^{\times})=0$.
My question is: How to show that $H^n(G, E^{\times})=0,\forall n\ge 2$ ?
When $G$ is a finite cyclic group, its cohomology is periodic: $H^{n+2}(G,M)\cong H^n(G,M)$ for all $n\ge1$. This proves your equation for odd $n$. But this fails for $n=0$. In this case $H^2(G,M)\cong{\hat H}^0(G,M)$, the $0$-th Tate cohomology group. This is defined to be $M^G/T(M)$ where $M^G=H^0(G,M)$ is the set of $G$-invariant elements of $M$ and $T(M)$ is the image of $M$ under the trace map $T:m\mapsto\sum_{g\in G}m^g$.
There are various ways of proving $\hat H^0(G,E^\times)\cong H^2(G,E^\times)$ is trivial. First, by direct computation: when $|F|=q$ and $|E|=q^n$, $E^\times$ is cyclic of order $q^n-1$, $(E^\times)^G$ has order $q-1$ and $T$ is powering by $(q^n-1)/(q-1)$. Second, by the theory of the Herbrand quotient, as $E^\times$ is a finite group, $|H^2(G,E^\times)|=|H^1(G,E^\times)|$. Third, $H^2(G,E^\times)$ is a subgroup of the Brauer group of $F$, which is trivial by Wedderburn's theorem on finite division rings.