Irreductible polynomial on a finite field of degree as large as wanted [duplicate]

A very simple counting estimation will show that such polynomials have to exist. Let $q=p^k$ and $F=\Bbb F_q$, then it is known that $X^{q^n}-X$ is the product of all irreducible monic polynomials over$~F$ of some degree$~d$ dividing $n$. The product$~P$ of all irreducible monic polynomials over$~F$ of degree strictly dividing $n$ then certainly divides the product over all strict divisors$~d$ of$~n$ of $X^{q^d}-X$ (all irreducible factors of$~P$ are present in the latter product at least once), so that one can estimate $$ \deg(P)\leq\sum_{d\mid n, d\neq n}\deg(X^{q^d}-X)\leq\sum_{i<n}q^i=\frac{q^n-1}{q-1}<q^n=\deg(X^{q^n}-X), $$ so that $P\neq X^{q^n}-X$, and $X^{q^n}-X$ has some irreducible factors of degree$~n$.

I should add that by starting with all $q^n$ monic polynomials of degree $n$ and using the inclusion-exclusion principle to account recursively for the reducible ones among them, one can find the exact number of irreducible polynomials over $F$ of degree $n$ to be $$ \frac1n\sum_{d\mid n}\mu(n/d)q^d, $$ which is a positive number by essentially the above argument (since all values of the Möbius function $\mu$ lie in $\{-1,0,1\}$ and $\mu(1)=1$). A quick search on this site did turn up this formula here and here, but I did not stumble upon an elementary and general proof not using anything about finite fields, although I gave one here for the particular case $n=2$. I might well have overlooked such a proof though.