TITLE: Explicit large finite fields in characteristic $2$
QUESTION [11 upvotes]: Every finite field of characteristic $2$ ist given by $\mathbb{F}_2[x]/P(x)$ for some irreducible polynomial $P\in \mathbb{F}_2[x]$.
For small degree, a simple algorithm gives a way to find $P$. Is there a way to give explicitely the polynomial $P$ for large degrees? I would be interested in a formula like
$$x^n+a_{n-1} x^{n-1}+\cdots +a_1 x+a_0$$
where $a_0,\ldots,a_{n-1}\in \mathbb{F}_2[n]$. This might be too much to ask for all degrees, so an infinite sequence of irreducible polynomials would be good.
EDIT: Some sequences have already be given. If you have more to share (with the proof of the irreducibility, either by giving the reference or explaining it), please go ahead.

REPLY [5 votes]: For $n=3^k$, the polynomial $p=x^{2n}+x^n+1$ is irreducible over $\mathbb{F}_2$.
Proof: By Rabin's irreducibilty test, it suffices to check that $p|x^{2^{2n}}-x$ and $\gcd(p,x^{2^{2n/3}}-x)=1$.
Note that the order of $x$ mod $p$ is $3n=3^{k+1}$. Hence, since $3^{k+1}|4^{n}-1$ by lifting-the-exponent lemma, we have $p|x^{2^{2n}-1}-1$.
Again by lifting-the-exponent lemma, we have $4^{3^{k-1}}-1=3^km$ for $m$ not divisible by $3$. Hence $x^{4^{3^{k-1}}-1}=x^{nm}\equiv x^n\  \text{or}\ x^{2n} \pmod{p}$ since $x^n$ has order $3$. As $(x^n-1)x^n=(x^{2n}-1)x^{2n}=1$, this mean that $x^{2^{2n/3}-1}-1$ is invertible in $\mathbb{F}_2[x]/(p)$. As $x$ is also invertible, we have shown that $\gcd(p,x^{2^{2n/3}}-x)=1$, as wanted.