$x^p - t$ has no roots in $\mathbb{F}_p(t)[x]$

Question

I am actually trying to prove that $x^p - t$ is irreducible in $\mathbb{F}_p(t)[x]$. My approach is to consider a splitting field $E$ so that: $$x^p - t = (x-\alpha_1)\dots(x-\alpha_p)=(x-\alpha)^p \ \ \ \ \text{for some $\alpha$ in $E$}$$ Then assuming by contradiction that $x^p - t = f(x)g(x)$, then $f(x) =(x-\alpha)^r$ with $r<p$. This implies that $\alpha \in \mathbb{F}_p(t)$. Should this be a contradiction right? Why can't $\alpha$ be here?

Answer 1

There is a duplicate of my question, but I'll post my own answer anyway: If $\alpha \in \mathbb F_p(t)$ then: $$ \alpha = \frac{F(t)}{G(t)} \ \ \text{where $F,G \in \mathbb F_p[x]$ } $$$$\implies \alpha^p = t = \frac{F(t^p)}{G(t^p)} $$ Then writing the coeffiecients for $F$ and $G$: $$ t(a_0 + ...+a_mt^{pm}) = b_0 + ... + b_nx^{np}$$ And then $t$ is algebraic over $\mathbb F_p$. Contradiction.