Is it true that $n!$ divides $p^n(p+1)(p^2+p+1)\cdots(p^{n-1}+\cdots+1)$?

Let $p$ be a prime number (though I suspect that this might be true for composite ones as well). Define $$f(p,n)=\frac{p^n(p+1)(p^2+p+1)\cdots(p^{n-1}+\cdots+1)}{n!}$$ where $n$ is a positive integer. Show that $f$ is always an integer.

I tested this for a handful of $n,p$ values and it seems to be true. I managed to show that for every $m\le n$, we can find a $k\le n$ such that $$1+p+\cdots+p^k\equiv 0\qquad(\text{mod }m)$$ But I am stuck and can't go further.

Edit

Thanks to all the nice answers. Now that it's clear the answer to the title is yes, we can furthur relax the $p^n$ term and replace it by $$p^{\lfloor n/(p-1)\rfloor}$$ so that it disappears when $p>n$.

Original version:

This is not true for $n=2p$. The numerator has only one factor that is a multiple of $p$, but the denominator is divisible by $p^2$.

Edited version:

This can be seen as follows. Let $q$ be a prime $\le n$. It is well known that the highest power $q^\ell$ that is a factor of $n!$ has exponent $$ \ell=\lfloor \frac nq\rfloor+\lfloor\frac n{q^2}\rfloor+\cdots $$ (this is a finite sum as the terms become zero after $q^k$ exceeds $n$). This is because $n!$ has $\lfloor\dfrac n q\rfloor$ factors that are divisible by $q$, $\lfloor\dfrac n {q^2}\rfloor$ factors that are (additionally) divisible by $q^2$ et cetera. We need to show that the numerator is divisible by $q^\ell$ as well.

If $q=p$ the first factor takes care of everything because $n\ge\ell$.

Otherwise, let $t$ be the order of $p$ modulo $q$, i.e. $t$ is the smallest positive integer such that $p^t\equiv1\pmod q$. It is known (Little Fermat) that $t\mid q-1$. Another standard exercise shows that $$ p^{tq^m}\equiv 1\pmod {q^{m+1}} $$ for all integers $m\geq0$.

Now we are well places to study an individual factor $$ x(r)=1+p+p^2+\cdots+p^{r-1}=\frac{p^r-1}{p-1}. $$

Assume next that $p\not\equiv1\pmod q$. It follows that $x(r)$ is divisible by the same power of $q$ as $p^r-1$. So $x(r)$ is divisible by $q$ whenever $r$ is a multiple of $q-1$, divisible by $q^2$ whenever $r$ is a multiple of $q(q-1)$ et cetera. It follows immediately that $\prod_{r=1}^n x(r)$ is divisible by $q^\ell$.

This leaves the case of $q\mid p-1$. In this case $p^k\equiv 1\pmod q$ for all $k$, so $x(r)\equiv r\pmod q$. In particular, $x(r)$ is divisible by $q$ whenever $q\mid r$. However, we need a little bit more to get some of the factors $x(r)$ divisible by higher powers of $q$ as well. A way forward is to observe that if $p\equiv1\pmod{q^k}$ for some $k\ge0$, then, for all $i$ we have that $$p^{q^i}\equiv1\pmod {q^{k+i}}$$ for all $i\ge0$. This forces $x(r)$ to be divisible by $q^i$ whenever $r$ is. This settles the last case.

No.

Take $p=5$. Then of the polynomials $p$, $p+1$, $p^2+p+1$, $\ldots$ only the polynomial $p$ is divisible by 5.

Yet $n!$ is divisible by $5^{\lfloor\frac{n}{5} \rfloor}$.

[In fact the above argument holds for any $p$ and $n$ large enough relative to $p$; indeed $n!$ is divisble by $p^{\lfloor\frac{n}{p} \rfloor}$]