https://www.youtube.com/watch?v=zP09Dw5D8nY
Here at precisely 31:00, he puts up a second claim, where $|A_d| \leq \phi(n)$where $A_d$ $=$ {$1 \leq a \leq p-1 | ord_p(a) = d$}, $d|p-1$ and $p$ is prime. (I understood the first one)
His proof is something which I am not able to understand, and I think he messed up somewhere.
His Claim:
Claim: If $A_d$ is not empty.
$|A_d| \leq \phi(n)$ for $A_d$ $=$ {$1 \leq a \leq p-1 | ord_p(a) = d$}, $d|p-1$ and $p$ is prime.
His Proof:
Take $f(x) \equiv x^d-1 \pmod{p}$ and $f(x) \equiv x^m-1 \pmod{p}$ where $m|d$.
He says that $A_d$ is the solutions of $g(x)$ removed from $f(x)$ which is understandable for $d$ being order.
Now he says that it will be $\phi(n)$ by defination.
This I don't understand, the words which he used to explain this statement are not clear to me, if you refer to the video you can understand what he is trying to say.
Question: What is he trying to do here?