TITLE: Another extension of Catalan: divisibility
QUESTION [11 upvotes]: The classical Catalan numbers are given by $C_n=\frac1{n+1}\binom{2n}n$, for which there is a plethora of generalizations of interpretations in the literature. Still, let us consider one more such:
$$C_n(k):=\binom{kn}{n,\dots,n}\prod_{j=0}^{k-1}\frac1{jn+1};$$
where the multinomial coefficient is simply $\frac{(kn)!}{n!^k}$.
Note. $C_n(2)=C_n$.
The following has been rediscovered many times and it is easy to prove: $C_n$ is odd iff $n=2^m-1$ for some $m$ (for example, see this paper by Postnikov and Sagan).
Such considerations prompted me to ask:

Question. Fix a prime $p$. Is this characterization valid?
  $$\text{$C_n(p)$ is not divisible by $p$ iff $n=1+p+\cdots+p^m$ for some $m$}.$$

UPDATE. Both solutions are great, in particular, I appreciated the variation of techniques that went into them.

REPLY [10 votes]: Here's a slightly different way to look at this. We have
$$C_n(k)=\prod_{j=1}^{k}\left(\frac{1}{(j-1)n+1}\binom{jn}{n}\right).$$
And each factor 
$$\frac{1}{(j-1)n+1}\binom{jn}{n}=\binom{jn}{n}-(j-1)\binom{jn}{n-1}$$
is an integer. For $C_n(p)$ let us look at the last factor $\frac{1}{(p-1)n+1}\binom{pn}{n}$. Since $\binom{pn}{n}$ is always divisible by $p$ we can say that if $C_n(p)$ is not divisible by $p$ then $p$ must divide $(p-1)n+1$, or in other words $n$ must be $1\pmod{p}$. If $n$ is written as $a_r\cdots a_1a_0$ in base $p$ then we have $a_0=1$ and by Lucas's theorem
$$\binom{pn}{n-1}\equiv \prod_{i=1}^{r}\binom{a_{i-1}}{a_{i}}\pmod{p}$$
Therefore in order for $\frac{1}{(p-1)n+1}\binom{pn}{n}$ not to be divisible by $p$ we must have $\binom{a_{i-1}}{a_{i}}\neq 0 \pmod{p}$. Therefore $1=a_0\geq a_1\geq \cdots \geq a_r$ which implies that all $a_i=1$. This is equivalent to saying $n=1+p+\cdots+p^r$ for some $r$. This also takes care of the other factors since for such $n$ we have $$\binom{jn}{n}\equiv j^r\neq 0 \pmod{p}.$$