The answer to your question is positive. Fix a prime $p$. Let $n$ be a positive integer. Let $T_1(n) = \binom{np}{n,\cdots,n}$, $T_2(n) = \prod_{j=0}^{p-1} (jn+1)$. You study $C_n(p)=\frac{T_1(n)}{T_2(n)}$.

**Case 1:** $p \mid n$. In this case, we have: $p \nmid T_2(n)$, since $jn+1 \equiv 1 \bmod p$. Moreover, $T_1(n) = \binom{np}{n} \binom{np-n}{n,\cdots,n} = \frac{np}{n} \binom{np-1}{n-1} \binom{np-n}{n, \cdots ,n}$ is divisible by $p$. Thus, $\frac{T_1(n)}{T_2(n)}$ is divisible by $p$.

Before the next main case, I prove a simple lemma.

**Lemma:** Let $p$ be a prime and let $a \in \{1,2,\cdots,p-1\}$. Then each $p$-adic digit of $-\frac{1}{a}$ is non-zero.

**Proof of Lemma:** If the $i$'th right-most digit is $0$, i.e. $$-\frac{1}{a} \equiv \overline{0 a_{i-2}\cdots a_0}_p \bmod {p^i},$$ then we have, after multiplying both sides by $a$,
$$\overline{(p-1)(p-1)\cdots (p-1)}_p \equiv a \cdot \overline{0 a_{i-2}\cdots a_0}_p \bmod {p^i},$$
and we reach a contradiction as the RHS is smaller than the LHS. $\blacksquare$

**Case 2:** $p \nmid n$. We compute the $p$-adic valuation of $T_1(n)$, $T_2(n)$. It is well known that $v_p(n!) = \frac{n-S_p(n)}{p-1}$ where $S_p(n)$ is the sum-of-digits of $n$ in base $p$. Thus
$$v_p(T_1(n)) = v_p((np)!) - p \cdot v_p(n!) = \frac{np-S_p(np)}{p-1} - p\frac{n-S_p(n)}{p-1} =$$
$$= \frac{p \cdot S_p(n) - S_p(np)}{p-1} = S_p(n).$$

Since $p\nmid n$, exactly one of the $p$ factors in $T_2(n)$ is divisible by $p$. Namely, if $n_0$ is the last digit of $n$ (i.e. $n\equiv n_0 \bmod p$, $n_0 \in \{1,2,\cdots, p-1\}$) and $m_0 \in \{ 1,\cdots,p-1\}$ satisfies $$n_0 m_0 = -1 \bmod p,$$ then
$$v_p(T_2(n)) = v_p(m_0 n+1).$$
Hence, the problem reduces to the following assertion:
$$S_p(n) \ge v_p (m_0 n+1), \text{ and equality holds iff }\exists k: n = \frac{p^k-1}{p-1}.$$
If $v_p(m_0 n+1)=j$ then $n \equiv \frac{-1}{m_0} \bmod {p^j}$. Let $y \in \{1,2,\cdots,p^j-1\}$ such that $y \equiv \frac{-1}{m_0} \bmod p^j$. We also denote it $\overline{\frac{-1}{m_0}}_{\bmod {p^j}}$. Since $S_p(n) \ge S_p( \overline{\frac{-1}{m_0}}_{\bmod {p^j}})$ with equality iff $n=\overline{\frac{-1}{m_0}}_{\bmod {p^j}}$, we have to prove that
$$(*) \forall m_0 \in \{1,2,\cdots, p-1\}: S_p( \overline{\frac{-1}{m_0}}_{\bmod {p^j}} ) \ge j, \text{ and equality holds iff }m_0=p-1.$$
(Note that the $p$-adic expansion of $\frac{-1}{p-1}$ is all ones, and in particular is the $p$-adic limit of $\frac{p^k-1}{p-1}$.)

To prove $(*)$ note that the last digit is $\overline{\frac{-1}{m_0}}_{\bmod {p}} \ge 1$ (equality for $m_0=p-1$) and that all $p$-adic digits of $\frac{-1}{m_0}$ are non-zero by the Lemma.