Divisibility of $a^x - 1$

Question

I'm currently working through a number theory script, and it's being taken as obvious that $x | y$ implies $a^x - 1 | a^y - 1$

I must be overlooking something, because I don't see how this is obvious. Could anyone please show me a short proof or nudge me into a direction where I'd be able to find one?

Answer 1

Hint:

$$a^k-1=(a-1)\sum_{i=0}^{k-1}a^i\tag{1}$$ What happens if you replace $a$ with $a^x$?

Also, don't feel sad about the "obvious" part. When a huge result is proven in physics it's a revolution. When the same thing happens in mathematics the result is obvious.

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Edit: As this question has been closed, I'll supply some details. Replacing $a$ with $a^x$ in $(1)$ yields $$a^{xk}-1=(a^x-1)\sum_{i=0}^{k-1}a^{xi}\tag{2}$$ Letting $y=xk$ we have $$a^x-1\mid a^y-1$$