The answer is: $\binom{k+N-1}{k}$ (binomial coefficient). Or $\binom{M+N-1}{M}$, since $M=k$ by your definition.
You questions is equivalent to the question: How many possibilites are there to distribute $k$ indistinquishable balls into $N$ pots?
Here is the argumentation:
$g =$ the number of possiblilities to distribute $N$ balls to $M$ pots, when the pots are not restriced in the number of balls they can hold.
$=$ number of possibilities to distribute $M-1$ walls and $N$ balls with in a long que of balls and walls (see illustration). (If the row starts with a wall, that means the first pot was empty.)
$=$ number of possibilities to distribute the $N$ balls to $N+M-1$ pots, which can hold one ball on maximum. (The previous walls are now represented by the empty pots.)
The last problem is the well-known case that is answered by the binomial coefficient $n+M-1\choose n$.
It might be a bit tough to follow the idea (and to understand my english), but maybe this picture will help you:
