How to prove $f(x^n)$ is divisible by $x^n-1$ The question states that if $f (x)$ is a polynomial such that $x-1|f(x^n)$ prove that $f(x^n)$ is divisible by $x^n-1$
This is how I proceeded since$x-1|f(x^n)$ 
$f(1)=0$ 
$\frac {f (x^n)-f(1)}{x^n-1}=g(x)$ 
since $f(1)=0$ 
$\frac{f(x^n)}{x^n-1}=g(x)$
 hence $x^n-1|f(x^n)$
I guess my proof is incorrect  can someone please  point out the mistakes and give the correct proof.
 A: José's solution is nice. And I agree with his assessment that an error you made is that you didn't explain why $g(x)$ is a polynomial. To see the error consider the following piece of faulty reasoning:
A False Fact. If $f(1)=0$ then $(x-1)^2\mid f(x)$.
Proof.
Write
$$
\frac{f(x)-f(1)}{(x-1)^2}=g(x).
$$
Because $f(1)=0$ we see that 
$$
\frac{f(x)}{(x-1)^2}=g(x).
$$
If your logic were valid, we could conclude that $(x-1)^2\mid f(x)$. But this result is clearly false in general. For example when $f(x)=x^2-1$ we have $f(1)=0$, but $(x-1)^2\nmid (x^2-1)$.

You can also do the following.
Let
$$
f(x)=a_0+a_1x+\cdots+a_mx^m,
$$
where $m$ is the degree of $f$. Then
$$
f(x^n)=a_0+a_1x^n+\cdots+a_mx^{nm}.
$$
We are given that $x-1\mid f(x^n)$. This means that $x=1$ is a zero of $f(x^n)$, or that $0=a_0+a_1+\cdots+a_m.$
But then
$$
\begin{aligned}
f(x^n)&=f(x^n)-f(1)\\
&=a_0(1-1)+a_1(x^n-1)+a_2(x^{2n}-1)+\cdots+a_m(x^{mn}-1).
\end{aligned}
$$
Here all the binomials $x^{kn}-1$, $k=0,1,\ldots,m$, are divisible by $x^n-1$. Hence so is $f(x^n)$.
A: $x-1|f(x^n)$ $\quad\Rightarrow\quad$ $f(1)=0$ $\quad\Rightarrow\quad$ $x-1|f(x)$ $\quad\Rightarrow\quad$ $x^n-1|f(x^n)$. 
A: Your proof is wrong because you did not prove that $g(x)$ is a polynomial.
Let $z\in\mathbb C$ be such that $z^n=1$. Then $f(z^n)=f(1)=0$. So, $z$ is a root of $f(x^n)$. Therefore, $f(x^n)$ is a multiple of$$(x-z_1)(x-z_2)\ldots(x-z_n),\tag{1}$$where $z_1,\ldots,z_n$ are the $n^\text{th}$ roots of unity. But $(1)=(x^n-1)$.
A: Assume the polynomial of degree $m$ has the roots $x_1,x_2,...,x_m$:
$$f(x)=a_0x^m+a_1x^{m-1}+\cdots +a_m=a_0(x-x_1)(x-x_2)\cdots (x-x_m).$$
Then:
$$f(x^n)=a_0(x^n-x_1)(x^n-x_2)\cdots (x^n-x_m).$$
Since $f(1)=f(1^n)$, then $x-1|f(x^n) \Rightarrow x-1|f(x)$. It implies that at least one root of $f(x)$ is $1$, hence $f(x^n)$ will have a factor $(x^n-1)$.
