No $p$-th root implies $X^{p^n}-a$ irreducible for all $n \in \mathbb{N}$

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While doing exercises of Chapter IV in Lang's algebra, I encountered the following problem:

Suppose char $K=p$. Let $a \in K$. If $a$ has no $p$-th root in $K$, show that $X^{p^n}-a$ is irreducible in $K[X]$ for all positive integers $n$.

However, one of the previous exercises states that

If $K$ is a finite field with $p^n$ elements, then every element of $K$ has a unique $p$-th root in $K$

My question is then if the two exercise are not contradictory. That is, if every element in $K$ has a unique $p$-the root, then it contradices the hypothesis of the first exercise and vice versa, if $a \in K$ has no $p$-th root, then not all element in $K$ has a $p$-th root.

Then what is it that I've missed? How is it possible that both statements are simoultaneously true?

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You have missed the fact not all fields of characteristic $p$ are of order $p^n$, for example $\mathbb{F}_p(t)$ is of infinite order.