Showing a function defined on a finite field is a bijection

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Let $p$ be a prime and $d \in \mathbb{N}$. If $(p-1,d)=1$ show that $f:F_p^x\to F_p^x$ given be $f(x)=x^d$.

When I first started this proof I considered two elements $a,b\in F_p^x$ where $a\neq b$. Then $f(a)=a^d$ and $f(b)=b^d.$ However, this being in a finite field, could it be the case that $a^d=b^d$? It's been a while since I have thought deeply about this topic, and now its bothering me that my answer wasn't satisfactory then or now.

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$(d,p-1)=1\implies$ that for any generator $g$ of the multiplicative group of the field, $f(g)=g^d$ also generates. Thus $f$ is surjective. Thus bijective, by finiteness.