<s>[INST] Prove or disprove $ p^{r+s}\mid q^{ke} - 1 \iff p^s \mid k$. Let $p$ be an odd prime and $q$ be a power of prime. Suppose $e := \min\{\, e \in \mathbb{N} : p \mid q^e - 1 \,\}$ exists. Put $r := \nu_p(q^e - 1)$ (that is, $p^r \mid q^e - 1$ and $p^{r+1} \nmid q^e - 1$). What I want to prove is the following:

$ \forall k \in \mathbb{N},~ \forall s \in \mathbb{N}_0,~p^{r+s} \mid q^{ke} - 1 \iff p^s \mid k.$

I feel that this is true (but perhaps some upper bounds may needed for $k$ and $s$). How can I prove or disprove this statement?

Example:
If $p = 3,~q = 5$ then $e = 2,~r = 1$.


*

*$5^{1 \cdot 2} - 1 = 2^3 \cdot 3^{1 + 0}$

*$5^{2 \cdot 2} - 1 = 2^4 \cdot 3^{1 + 0} \cdot 13$

*$5^{3 \cdot 2} - 1 = 2^3 \cdot 3^{1 + 1} \cdot 7 \cdot 31$

*$5^{4 \cdot 2} - 1 = 2^5 \cdot 3^{1 + 0} \cdot 13 \cdot 313$

*$5^{5 \cdot 2} - 1 = 2^3 \cdot 3^{1 + 0} \cdot 11 \cdot 71 \cdot 521$



My attempt:
In case $(\Leftarrow)$,
$$\begin{align}
q^{ke} - 1 &= ((q^e - 1) + 1)^k - 1 \\
&= (q^e - 1)^k + k(q^e - 1)^{k - 1} + \dotsb + \binom{k}{2}(q^e - 1)^2 + k(q^e - 1).
\end{align}$$
I can see that the last term is divisible by $p^{r + s}$ from the assumption. But is it true for the former terms?
In case $(\Rightarrow)$, we may start by introducing the quotient and the remainder $k = ap^s + b~(0 \le b < p^s)$ and calculate similar expansion as before. 
 [/INST] Let $x=q^e-1$.  We have $q^{ke}-1 = \exp(k\cdot\ln (1 + x))-1$.  Because $\nu_p (x)\geq 1$, we have $\nu_p (\ln (1 + x)) = \nu_p(x)\geq 1$ (look at power series), so that $k\cdot \ln(1+x)$ is in the radius of convergence of the $p$-adic exponential (here we use the fact that $p$ is odd).
Looking again at the power series, we have $\nu_p(q^{ke}-1) = \nu_p (\exp(k\cdot\ln (1 + x))-1) = \nu_p(k\cdot\ln (1 + x)) = \nu_p(k) + \nu_p(q^e-1)$, which is exactly what we wanted to show.
</s>