Is there an "easy" proof of the existence of a primitive element in a finite field? My book uses three lemmas and a theorem to get there, and I can't shake the feeling that there's an easier way of doing it.

I've made some attempts using proof by smallest counter-example, looking at the set of all finite fields with a certain number of elements. E.g. it's trivially true for the set of all finite fields with 2 elements, and then moving from e.g. $A_k$ to $A_{k+1}$ (where $A_k$ is the set of all finite fields with $k$ elements) you pick a field $F\in A_{k+1}$ which does not have a primitive element, and look at a subfield of it where you exclude one element. This subfield then does have this property. After that I got stuck.

Anyone know of any relatively "clean" proofs?

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    $\begingroup$ Are you sure by removing one element of a field you get a subfield? $\endgroup$ – coffeemath Jan 9 at 15:42
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    $\begingroup$ See Andrea's answer at Finite subgroups of the multiplicative group of a field are cyclic. In particular, the whole multiplicative group of non-zero elements is cyclic, so there is a primitive element. $\endgroup$ – Parcly Taxel Jan 9 at 15:59
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    $\begingroup$ Depends. If you have covered the structure theorem of finitely generated (in particular finite) abelian groups, then it becomes a bit easier. $\endgroup$ – Jyrki Lahtonen Jan 10 at 4:47

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