For any prime number $p$ and $m \in \Bbb N$ there exists a field of cardinality $p^m.$ I am studying Galois theory from an NPTEL online lecture series.  While watching the lecture on finite fields I came across a theorem which I already stated in the title. The proof of this result went as follows $:$
Let us consider the polynomial $X^{p^m} - X$ in $\Bbb Z_p [X].$ Then Kronecker's theorem guarantees that there exists a finite field extension of $\Bbb Z_p$ say $L'$ such that the above polynomial splits completely into linear factors in $L'[X].$
Let $V_{L'} \left ( X^{p^m} - X \right )$ be zero set of the polynomial $X^{p^m} - X$ in $L'.$ Then clearly $\left | V_{L'} \left (X^{p^m} - X \right ) \right | = p^m.$ Let $L = V_{L'} \left (X^{p^m} - X \right ).$ If we can prove that $L$ is a subfield of $L'$ then we are through.
This is where I got stuck. I have proved almost all the properties which will make $L$ a subfield of $L'$ except the existence of additive inverse of every element in $L.$ For showing this property I take an element $x \in L.$ I need to show that $-x \in L.$ Since $x \in L$ we have $x^{p^m} - x = 0.$ Now let $p$ be an odd prime \begin{align*} (-x)^{p^m} - (-x) & = -x^{p^m} + x\\ & = - \left (x^{p^m} - 1 \right )\\ & = 0. \end{align*}
But if $p=2$ then I cannot say that. Because then $(-x)^{2^n} = x^{2^n}.$ So $(-x)^{2^n} - (-x) = x^{2^n} + x$ which may not be zero in $L.$ 
So my question is $:$ How do I prove the result for $p=2$ i.e. how do I prove that for any $m \in \Bbb N$ there exists a field of cardinality $2^m$? Any help regarding this will be highly appreciated.
Thank you very much.

This is very simple. If $p=2$ then we have $1=-1$, right? So in that case:
$(-x)^{2^n}=(-1)^{2^n}x^{2^n}=1\times x^{2^n}=1\times x=(-1)\times x=-x$
