Deducing that $a$ has order $n$ in a group $G$. Suppose that $a^n = 1$ in a group $G$. Prove that if, for every prime $p$ dividing $n$ we have $a^{n/p}\ne 1$, then $a$ has order $n$.

For the sake of contradiction, suppose $r<n$ is the order of $a$. Then $r$ is a divisor of $n$, and there exists a prime $p$ which divides $n/r$. Putting this together, $$a^{n/p}=a^{(rn/r)/p}=a^{rs}=(a^r)^s=1,$$ where $s$ is the integer $n/rp$.
