Prove that exactly $\varphi (d)$ residues have order $d$ modulo $p$

Question

Prove that exactly $\varphi (d)$ residues have order $d$ modulo $p$ for all $d \mid \varphi(p)$ where $p$ is prime.

I was wondering if there was a way to solve this using number theory and not group theory. Here was the start of my attempt:

Let $g$ be a primitive root modulo $p$, so that $\text{ord}_p(g) = \varphi(p) = p-1$ and $\{g,g^2,\ldots,g^{p-1}\}$ is the complete residue system modulo $p$. It suffices to find the number of $m$ such that $\text{ord}_p(g^m) = d$.

We know that $g^{dm} = g^{(p-1) \cdot \frac{dm}{p-1}}$.

How do we continue from here?

Answer 1

If $d \mid \phi(p)$ then write $d'$ for the number such that $dd' = \phi(p)$. Now $h = g^{d'}$ has order $d$ (if $h^k = 1$ for $k