abstract algebra - On divisors of $p^n-1$

This question raised from discussions around my previous question. This may seem trivial or easy, but I am so confused and can't see the answer. So I will be so grateful if you would help me please.

For the natural number $n$ and prime number $p$, it is likely possible that there exists a polynomial $f(x)\in\mathbb{Z}[x]$ of the form $f(x)=\prod_{i=1}^{m}(x^{\lambda_i}-1)$ such that $deg(f)>n$ and $f(p)\mid p^n-1$. For example, take $p=3$, $n=2$ and $f(x)=(x-1)^3$ we have $(3^1-1)(3^1-1)(3^1-1)\mid 3^2-1$. However, with assumptions $p\geq 3$ and $n\geq 4$, I could not find any such polynomial $f$ with $\deg(f)>n$ and $f(p)\mid p^n-1$. So it made me to claim that the following statement might be true:

Let $p\geq 3$ be a prime number and $n\geq 4$. Then for every $f(x)\in\mathbb{Z}[x]$ of the form $f(x)=\prod_{i=1}^{m}(x^{\lambda_i}-1)$ such that $\deg(f)>n$, we have $f(p)\not\mid p^n-1$.

Is the above assertion true? If yes, would you please hint me how to prove it (or refer me to a reference)?
Thank you in advance.