No, that expression isn’t correct. The correct expression is
$\binom{N+M-1}{M-1}=\binom{N+M-1}N=\frac{(N+M-1)!}{N!(M-1)!}\;;$
this formula is derived here. As André notes in the comments, this can be calculated without division, by using the identity
$\binom{n}m=\binom{n-1}m+\binom{n-1}{m-1}$
with the initial conditions $\binom{n}0=1$ for all integers $n\ge 0$ and $\binom0m=0$ for all integers $m>0$.
Added: To see why your reasoning doesn’t work, consider the case of $3$ balls and $2$ buckets, labelled $A$ and $B$. Suppose that we put the balls into the buckets one at a time; then the $2^3=8$ possibilities are $AAA,AAB,ABA,ABB,BAA,BAB,BBA$, and $BBB$. One of these puts $3$ balls in $A$ and none in $B$; $3$ of them put $2$ balls in $A$ and $1$ in $B$; $3$ of them put $1$ in $A$ and $2$ in $B$; and $1$ of them puts all $3$ balls in $B$. In other words, the number of permutations corresponding to each distribution of the unlabelled balls is not constant: it depends on the distribution. Here the count of $2^3$ sequences counts two of the possible distributions correctly, but it overcounts the other two by a factor of $3$. And this variation only gets worse as the numbers of balls and buckets go up.