Let $p$ be a prime. Try to show that $[\mathbb F_p (t): \mathbb F_p(t^p)] = p$.
I tried to search for relevant hints and come across with these posts.
Example for non separable elements?
Non-separable, infinite field extensions of non-zero characteristic
I am not sure why the minimal polynomial is $X^p - t^p = (X - t)^p$ and how minimal polynomial helps here.
I know that $\mathbb Z_p$ has characteristic $p$, which gives us for all $a, b \in \mathbb{F}_p$: $$(a + b)^p = a^p + b^p$$
But isn't $t$ here a formal variable?
Also, I know a similar theorem which states that if $p(x)$ is the minimal polynpmial of $\alpha$ over $F$, we have $[F(\alpha) : F] = \deg(p)$.
But the $\mathbb F_p (t)$ and $\mathbb F_p(t^p)$ does not quite fit the above theorem.
Let $K=\mathbb{F}_p(t^p)$ and let $f(X)=X^p-t^p\in K[X]$. Note that $t$ is irreducible in $\mathbb{F}_p[t^p]$. Since $\mathbb{F}_p[t^p]$ is a PID, it follows that $(t^p)$ is a prime ideal, $1\not\in (t^p), t^p\not\in t^{2p}$ and so by Eisentstein's Criteria $X^p-t^p$ is irreducible in $\mathbb{F}_p[t^p][X]$. Then by Gauss's lemma it is irreducible in $\mathbb{F}_p(t^p)[X]=K[X]$. Hence $L = K[X]/(X^p-t^p)$ is an extension of degree $p$. As you pointed out, $X^p-t^p=(X-t)^p$ so $t$ is the only root of $f$. Therefore $t\in L$ and so $L=K(t)=\mathbb{F}_p(t)$.