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Let $P(x)$ be a polynomial with real coefficients of degree $ninmathbb{N}$. I want to prove that if $(x-1)|P(x^n)$ then $(x^n-1)|P(x^n)$.

the asertion is that if $x=1$ satisfies $Q(x^n)=0$, then so does $x=zeta$ where $zeta^n = 1$.

set:
$$
Q(x^n) = sum_{k=0}^M a_k (x^n)^k
$$
since $x^n =1$ is a root, we also have:
$$
0 = sum_{k=0}^M a_k
$$
subtracting the second equation from the first:
$$
Q(x^n) = sum_{k=1}^M a_k(x^{kn} - 1)
$$

clearly, if $zeta^n$ = 1 then $zeta^{kn}=1$, so $x=zeta$ is a root of $Q(x^n)=0$

A short but cute exercise. Recall the following

Bezout's theorem (on divisibility): for any commutative ring $A$ , element $a in A$ and polynomial $f in A[X]$ it holds that $X-a | f$ if and only if $f(a)=0_A$.

Let us apply this to your case. Allow me to redenote the polynomial $P$ by $f$ (matter of taste) and to introduce $g=f(X^n)$. As $X-1 | g$ we can infer that $g(1)=f(1)=0$, hence we also have $X-1 | f$. Writing this relation as $f=(X-1)h$, for a certain $h in mathbb{R}[X]$, you can employ the universality property of polynomial rings to substitute $X^n$ for $X$ in the above factorisation of $f$ and obtain
$$g=(X^n-1)h(X^n)$$

which means $X^n-1 | g$, q.e.d.

$(x-1)|Q(x^n)=P(x)implies P(1)=0$ by the Factor Theorem.

$implies P(1)=Q(1^n)=Q(1)=0$

$implies (x-1)|Q(x)$

$implies (x^n-1)|Q(x^n)$