Prove that $n$ divides $\phi(a^n -1)$ where $a, n$ are positive integer without using concepts of abstract algebra I need to show that $n$ divides $\phi(a^n -1)$ where $a, n$ are positive integer without using concepts of abstract algebra

I know that $$a^n\equiv 1\pmod {a^n-1}$$

How do I proceed from there?
In general (for $\gcd (a,m)=1)$, if $a_k≡1 \mod m $, then $ord_m(a) ~∣~k.$ In particular, since $a^{ϕ(m)}≡1(\mod m)$, we have $ord_m(a)∣ϕ(m)$. We have $ord_{(a^n−1)}a=n.$

Proof: Let there exist $d&lt; n$ such that $ a^d ≡ 1 \mod(a^n-1) $ .

=> $ (a^n-1)~ |~ (a^d -1)$ which is not possible since $d &lt; n $.

=> $ord _{a^n−1} =n.$

=> $n~|~ \phi(a^n-1)$
The order of the multiplicative group mod $a^n-1$ is $\phi(a^n-1),$ which (together with your start) proves your statement, but I assume that since the word "group" is used, this is not the proof asked for.