Q:

Group theory: counting the number of elements in $\mathbb{Z} _p ^*$

Let $p$ be a prime number.
Let $d$ is a divisor of $(p-1)$
Let $G$ be a group of integers $\{1,2,\cdots,p-1\}$ under multiplication modulo $p$. 
How may one prove that the number of elements $a$ in $G$ such that $a^d\equiv1$ (mod $p$) is exactly $d$?
Given $X^{p-1}-1=(X^d-1)(X^{d(k-1}+X^{d(k-2)}+\cdots +X^d+1)$ where $k=\frac{p-1}{d}$

A:

Hints:
Over a field (i.e., with coefficients from a field), a polynomial of degree $\;d\;$ has at most $\;d\;$ different roots.
Look now at $\;x^d-1\in\Bbb F_p[x]\;,\;\;with\;\;\Bbb F_p:=\Bbb Z/p\Bbb Z$

