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If $a$ is any integer and $n$ is divisible by $d$, then $a^n-1$ is divisible by $a^d-1$.

Is it valid to solve this using Fermat's? Saying that since $d|n$, $n=ad$ for some integer $a$. Then $a^n-1=a^{ad}-1$ and $a^d \equiv 1 (\mod d+1)$ so $a^{ad}-1 \equiv 0 (\mod d+1)$. Also, $a^d-1\equiv 0 (\mod d+1)$. Is this valid for the question? I am hesitant because I used the specific scenario of $\mod d+1$. Also does them both being $0$ mean that they divide each other... Thank you.

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    Have you tried just doing long division? – 2017-02-25
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    You can't assume $n=ad$ for the *same* $a$. – 2017-02-25
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    http://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 and – 2017-02-26

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Hint: $a^n-1=a^{kd}-1=(a^d)^k-1=(a^d-1)\left((a^d)^{k-1}+(a^d)^{k-2}+\cdots +a^d+1\right)\,$.

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    Thank you. I think I definitely went down the wrong path. – 2017-02-25
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    To answer a now deleted comment, the above uses the [difference of powers identity](http://artofproblemsolving.com/wiki/index.php/Sum_and_difference_of_powers): $$a^n-b^n=(a-b)(a^{n-1} + a^{n-2}b+a^{n-3} b^2+\cdots+a b^{n-2}+b^{n-1})$$ – 2017-02-25