Where $d \in \mathbb{Z}_p \setminus \{ 0\}$.
I have two cases to consider: when $n \mid p$ and when $n \nmid p$.
For $n \nmid p$ I found that there are $n$ distinct roots in its splitting field, $K$, because of the relation that $$f'(x) \neq 0 \iff f(x) \text{is separable} $$
And since $n \nmid p$, $n \not \equiv 0 \ (mod \ p)$ so $f'(x)$ is not identically $0.$ I was wondering if this is correct?
For $n \mid p$ i found this, but feel uncomfortable because I haven't been exposed to polynomial rings of the form like $F[X^p]$ before, nor understand the significance of $a \neq b \implies a^p \neq b^p$ in the solution. Also, it describes an upper bound for distincy roots, whereas the question I'm working on says to give a precise expression for the number of distinct roots in its splitting field. I also don't understand the significance of the condition that $d \neq 0$ in my question either. Any guidance is appreciated whether on a new approach or to clarify the linked one.
Thanks for reading!
This is wrong. This equivalence only holds when $f$ is irreducible, which your $f$ may not be. The correct equivalence is that $f$ is separable iff $f$ and $f'$ are relatively prime (since common roots of $f$ and $f'$ are the same as repeated roots of $f$). See where that gets you for your particular $f$ (here you will need to use the assumption that $d\neq 0$.
In the case $p\mid n$, I don't have anything better to say than what Jyrki Lahtonen commented, so I'll just repeat his idea here. The point is that if $n=pm$, then $x^{n}-d=(x^m)^p-d$ factors as $(x^m-d^{1/p})^p$ (where in fact $d^{1/p}=d$ since $d\in\mathbb{Z}_p$, but that's not important). So, the distinct roots of $x^n-d$ are the same as the distinct roots of $x^m-d^{1/p}$. Now if $m$ is still divisible by $p$ you can repeat this process, until you've reduced to the case where $p$ does not divide the degree and you're in the first case.