Square root in $\mathbb F_{2^n}$

Question

Let $\mathbb F_{2^n}$ be a finite field with $2^n$ elements.

I am just wondering if it is true that for all $n\in \mathbb N$ all elements of $\mathbb F_{2^n}$ have square roots, i.e for all $a\in \mathbb F_{2^n}$ there is an element $b\in \mathbb F_{2^n}$, with $a=b^2$?

It may be very easy but I can not see now how to see it.

Answer 1

Let $x,y\in \Bbb F_{2^n}$ be two elements with $x^2=y^2\implies (x-y)(x+y)=0$, since a field is an integral domain we have either $x=y$ or $x=-y=y$ as $\text{char}(\Bbb F_{2^n})=2$. Hence, $x\not=y\iff x^2\not=y^2$, so that $\big\{x^2|x\in \Bbb F_{2^n}\big\}=\Bbb F_{2^n}$. So every element is a square root.

Answer 2

Do u know that for every $a\in \mathbb F_{2^n}$ $a^{2^n}-a=0$.