I am 16. I am a new french member and I hope me to improve my standart in english.

a, m and n are three natural entires.
I must compare :

$\displaystyle (a^m-1\wedge a^n-1)$ and $\displaystyle a^{m\wedge n}-1$
what that I do :

I have posed: $\displaystyle d=m\wedge n$

$\displaystyle m=dq$ et $\displaystyle n=dk$ with $\displaystyle q\wedge k=1$.

$\displaystyle \begin{cases} a^n-1= (a^d-1)\sum_{i=0}^{k-1}(a^d)^{k-1-i} \\ a^m-1= (a^d-1)\sum_{i=0}^{q-1}(a^d)^{q-1-i}\end{cases}$


$\displaystyle a^n-1\wedge a^m-1 = (a^d-1)\times (\sum_{i=0}^{q-1}(a^d)^{q-1-i} \wedge \sum_{i=0}^{k-1}(a^d)^{k-1-i} )$

so here I suppose that: $\displaystyle a^n-1\wedge a^m-1= a^{m\wedge n}-1$

and now I try to show:

$\displaystyle (\sum_{i=0}^{q-1}(a^d)^{q-1-i} \wedge \sum_{i=0}^{k-1}(a^d)^{k-1-i})=1 $

but I am not finding