Existence of primitive roots modulo a prime

Statement

Let $p$ be a prime number. Then there is a primitive root modulo $p$. That is, the multiplicative group $(\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic of order $p - 1$.

Remarks

A proof: a polynomial of degree $d$ has at most $d$ roots in the field $\mathbb{Z}/p\mathbb{Z}$. So for each $d \mid p - 1$, at most $\varphi(d)$ elements have order $d$. The sum of the totient over the divisors then forces exactly $\varphi(d)$ elements of each order $d$. In particular, there are exactly $\varphi(p - 1)$ primitive roots modulo $p$ in $\{1, \dots, p - 1\}$. More generally, each finite subgroup of the multiplicative group of a field is cyclic.

Explanations

L1 human

For each prime $p$, some number $g$ has powers that run through all of $1, 2, \dots, p - 1$ modulo $p$. For $p = 11$, the number $2$ works: its powers are $2, 4, 8, 5, 10, 9, 7, 3, 6, 1$.