How to approach this divisibility problem on polynomial Prove that the polynomial $x^{9999} + x^{8888} + x^{7777} + ... + x^{1111} + 1$ is divisible by $x^9 + x^8 + x^7 + ... + x + 1$.
I have no idea how to approach this problem, some help would be appreciated.

$$ ( x^9 + x^8 + \ldots + x + 1) \left(x^{9990} - x^{9989} \right) =  x^{9999} - x^{9989}$$
So,
$$ x^{9999} - x^{9979} = x^{9999} - x^{9989} + x^{9989} - x^{9979} =  ( x^9 + x^8 + \ldots + x + 1) \left(x^{9990} - x^{9989} + x^{9980} - x^{9979} \right) $$
Continuing in this way, we see that:
$$ x^9 + x^8 + \ldots + x + 1\ \vert\ x^{9999} - x^9, $$
$$ x^9 + x^8 + \ldots + x + 1\ \vert\ x^{8888} - x^8, $$
$$ \ldots$$
$$ x^9 + x^8 + \ldots + x + 1\ \vert\ x^{1111} - x^1, $$
$$ x^9 + x^8 + \ldots + x + 1\ \vert\ 1 - 1, $$
Therefore,
$$ x^9 + x^8 + \ldots + x + 1\ \vert\ x^{9999} - x^9 + x^{8888} - x^8 + \ldots + x^{1111} - x^1 + 1 - 1,\quad \text{ i.e.,} $$
$$ x^9 + x^8 + \ldots + x + 1\ \vert\ x^{9999} + x^{8888} + \ldots + x^{1111} + 1 - \left( x^9 + x^8 + \ldots + x + 1 \right) $$
$$ \implies x^9 + x^8 + \ldots + x + 1\ \vert\ x^{9999} + x^{8888} + \ldots + x^{1111} + 1.$$
