Suppose that $f(x) \in \Bbb Z_p[x]$ and is irreducible over $\Bbb Z_p$, where $p$ is prime. If $\deg f(x) = n$, prove that $\Bbb Z_p[x] / \langle f(x) \rangle$ is a field with $p^n$ elements.
I can see that $\Bbb Z_p[x] / \langle f(x) \rangle$ is a field because $\Bbb Z_p$ is a field and $f(x)$ is irreducible, but I can't figure a way to show that it has $p^n$ elements.
I know I can write any $g(x) + \langle f(x) \rangle$ as $q(x)f(x) + r(x) + \langle f(x) \rangle$ where $r(x)$ is the remainder upon division, but I can't seem to figure what the remainder would be when dividing arbitrary elements.
Regarding the $p^n$ elements in the field $\mathbb{Z}_p [x] / \langle f(x) \rangle$, note that any $q \in \mathbb{Z}_p [x] / \langle f(x) \rangle$ has the form $q(x) = \sum_{i = 0}^{n - 1} \alpha_i x^i$, where $\alpha_i \in \mathbb{Z}_p$ for any $0 \leq i \leq n -1$. From this it follows that there are $p^n$ different polynomials.
That $\mathbb{Z}_p [x] / \langle f(x) \rangle$ has $p^n$ elements also immediately follows from the fact that $\mathbb{F}_{p^n} \cong \mathbb{Z}_p [x] / \langle f(x) \rangle$, where $\mathbb{F}_{p^n}$ denotes the Galois field. Of course you should be aware with this isomorphism. Therefore, the first approach is more straightforward.