I'm trying to find the splitting field of $x^{p^{n}}-1\in\mathbb{Z}_{p}\left[x\right]$ where $p$ is a prime and integer $n\geq1$.
If $\alpha_{1},\dots,\alpha_{p^{n}}$ are the roots then $\mathbb{Z}_{p}\left(\alpha_{1},\dots,\alpha_{p^{n}}\right)$ is the splitting field, but what do elements of this field look like exactly?
I've factored $x^{p^{n}}-1$ as follows but I'm not sure what else to do $$\begin{align*} x^{p^{n}}-1 & =\left(x-1\right)\frac{x^{p^{n}}-1}{x-1}\\ & =\left(x-1\right)\prod_{i=0}^{n-1}\frac{x^{p^{i+1}}-1}{x^{p^{i}}-1}\\ & =\left(x-1\right)\prod_{i=0}^{n-1}\sum_{j=0}^{p-1}x^{p^{i}j} \end{align*}$$
I think you mean $x^{p^n}-x$. A field $F$ of characteristic $p$ satisfies $|F|=p^k$ for some integer $k$, so $|F^×|=|F \setminus \{0\}|$ $=p^k-1$. Thus as $p$ does not divide $p^k-1$ for any positive integer $k$, it follows by Lagrange's Thm that $|F^×|$ cannot have any group closed under multipication of order that is a multiple of $p$.
In a field $F$ of size $p^n$, all $p^n$ elements $\alpha \in F$ satisfy $\alpha^{p^n}-\alpha = 0$. So as the number of roots of $p(x)=x^{p^n}-x$ that are in $F$ is deg$(p)$, it follows that the polynomial $p(x)=x^{p^n}-x$ splits in $F$.
Meanwhile in characteristic $p$ note that $(x-1)^{p^n}=x^{p^n}-1$. So $1$ is a root w multiplicity $p^n$; i.e., $x^{p^n}-1$ factors completely in $\mathbb{F}_p[x]$ to $(x-1)^{p^n}$. And thus the splitting field is simply the $p$-element field $\mathbb{F}_p$.