I am currently trying to find a primitive element of the multiplicative group of field $GF(p)$. Since the numbers are relatively small, I know the factorization of
$$\phi(p)=p-1 = {p_1}^{k_1} {p_2}^{k_2} ... {p_n}^{k_n}$$
Wikipedia says that $m\in GF(p)$ is a generator if
$$m^{\phi(n)/p_k} \not\equiv 1 \mod p \quad \forall k=1..n$$
However, there is no clear explanation why it is the case. Could you please help me to understand this?
Can you see that every proper divisor of $\phi(p)$ is a divisor of $\phi(p)/p_k$ for some $k$? So if the order of $m$ isn't $\phi(p)$, then it's a divisor of some $\phi(p)/p_k$?