I found the answer in this document named Text classification and Naive Bayes, I think it gives detailed explanations for you questions.
Of course, I would like to share some of my own explanations, too.
Firstly, I need to introduce some notations, this is important to correct your statements about the question:
The naive bayes assumption shows:
$P(X|Y=c_k)=P(X_1,X_2,...,X_n|Y=c_k)=\prod_{i=1}^{n}P(X_i|Y=c_k)$, here the bold character $X_i$ represent r.v, not the values.
The key of multinomial NB is that it assumes $P(X_i|Y=c_k)$ is a multinomial distribution with $n=1$, especially $P(X_i|P=c_k)$ obeys the same distributions for all $X_i$ (actually this is the simplified situation, but it is common for text classification).
Suppose $X_i$ chooses values from the set $\{a_1,a_2,...,a_m\}$,let $p_j$ denotes the probability of $P(X_i=a_j|Y=c_k)$, $n_j$ denotes the occurrences of $X_i=a_j$ and $n=n_1+n_2+...+n_m$,(here $n$ is nothing to do with $n$ in $X_n$, just for convenience).
So $P(X_i|Y=c_k)=P((n_1,n_2,...n_m)|Y=c_k)=\frac{n!}{n_1!n_2!···n_m!}p_1^{n_1}p_2^{n_2}···p_m^{n_m}=p_1^{n_1}p_2^{n_2}···p_m^{n_m}$, because $n=1$.
Now we turn to the joint probability $P(X_1,X_2,...,X_n|Y=c_k)$.Since every $X_i$ choose values from $\{a_1,a_2,...,a_m\}$, the result sequence $(X_1,X_2,...,X_n)$ can be represent as $(n_1,n_2,...,n_m)$, which means the occurrences of different results $\{a_1,a_2,...,a_m\}$ , actually this is a multinomial distribution with repeats of $n$. We have $P(X_1,X_2,...,X_n|Y=c_k)=P((n_1,n_2,...n_m)|Y=c_k)=\frac{n!}{n_1!n_2!···n_m!}p_1^{n_1}p_2^{n_2}···p_m^{n_m}$, here $n=n_1+n_2+...+n_m$.
Well, you can see for the present, $\prod_{i=1}^{n}P(X_i|Y=c_k)$ is not equal to $P(X_1,X_2,...,X_n|Y=c_k)$ because they are different in the coefficient $\frac{n!}{n_1!n_2!···n_m!}$.
Here comes a another assumption: positional independence(see the document linked above),it assumes the positions of result $a_j$ in the sequence $(X_1,X_2,...,X_n)$ doesn't matter. This assumption means given an occurrences of $a_j$,say $n_j$, the different combinations $C_n^{n_j}$ doesn't matter. For the coefficient in multinomial distribution, $\frac{n!}{n_1!n_2!···n_m!}=C_n^{n_1}C_{n-n_1}^{n_2}...C_{n-n_1-n_2...-n_{m-1}}^{n_m}$,namely this coefficient is eliminated by the positional independence assumption.
Maybe the way I interpret the feature vector $(X_1,X_2,...,X_n)$ is some kind of confusing, so I strongly recommend you reading the document I linked because you can get a better understand in the specific text classification problem.