Each element is a square of some element I have to show that each element of $\mathbb{F}_{2^n}$ is a square of some element.
Could you give me some hints how I could do that??
 A: In $\mathbb{F}_{q^n}$  every element $x$ satisfies $x^{q^n}=x$. In fact, for $x=0$ this is clear, and for $x \neq 0$ it follows from Lagrange's theorem applied to the multiplicative group $\mathbb{F}_{q^n}^\times$. In particular, every element of $\mathbb{F}_{2^n}$ is a square.
A: Consider the map $\phi : \mathbb{F}_{2^n} \to \mathbb{F}_{2^n}$ given by $\phi(x) = x^2$. You can show that this is a homomorphism between rings (use the fact that $\operatorname{char}(\mathbb{F}_{2^n}) = 2$), so its kernel is an ideal in $\mathbb{F}_{2^n}$. As $\mathbb{F}_{2^n}$ is a field, $\ker\phi = \{0\}$ or $\ker\phi = \mathbb{F}_{2^n}$, but you should be able to rule out the latter by demonstrating an element in $\mathbb{F}_{2^n}$ which has non-zero square. Once you have done this, you see that $\phi$ is an isomorphism; in particular, it is surjective.
A: Consider the map $\phi : \mathbb{F}_{2^n} \to \mathbb{F}_{2^n}$ given by $\phi(x) = x^2$. 
Then, $\phi$ is injective because $x^2=y^2$ iff $0=x^2-y^2=(x+y)(x-y)=(x-y)^2$. Since there are no zero divisors in a field, $x-y=0$ and so $x=y$.
Since the field is finite, $\phi$ is surjective and every element is a square.
