In the book "A Classical Introduction to Modern Number Theory", I saw the following theorem (p. 43):
If $p\neq 2$, and $p\nmid a$ then $p^{l-1}$ is the order of $(1+ap)$ mod $p^l.$
i.e. $(1+ap)^{p^{l-1}}\equiv 1(\text{mod } p^{l})$, and $(1+ap)^{p^k}\not\equiv 1(\text{mod } p^l)$ for $0< k< p^{l-1}.$
The question is:
Suppose we consider the finite field $\mathbb{F}_{p^l}$ , $p\neq 2$, and take $a=1$. Then by above theorem, the multiplicative order of $(1+p)$ in $\mathbb{F}_{p^l}$ is $p^{l-1}$. But, as $1+p$ is non-zero element of the Field $\mathbb{F}_{p^l}$, so it is an element of the cyclic group $\mathbb{F}^*_{p^l}$ of order $p^l-1$, so its multiplicative order should divide $p^l-1$, and hence can not be prime power, hence the order can not be $p^{l-1}.$
What's going wrong here?