Is there a trick to compute this multinomial-looking sum? The series I want to sum has this form
$\displaystyle \sum_{} 1^{l_1}(1+c)^{l_2} (1+2c)^{l_3} \cdot \ldots \cdot  (1+(N-1)c)^{l_{N}}$
for some constant $c$ and positive integers $N$ and $L$. Here the sum is taken over all ordered (finite) sequences $(l_1,l_2, \ldots , l_N)$ that satisfy $l_1+l_2+ \ldots l_N = L$. There are exactly $ {L+N-1} \choose {N-1}$ such sequences and the same number of terms to sum.
The reason this looks like the multinomial formula is because it came out of something similar. In particular it came out of the similar formula for the order-$L$ derivative of a product of $N$ many functions. The only difference is those formulas have coefficients that depend on $(l_1,l_2, \ldots , l_N)$.
So this guy is screaming out that he wants to be recovered as a coefficient of something. For example if I were to add up all the terms $1,1+c,1+2c,\ldots, 1+(N-1)c$ I'd get $\frac{n(n-1)}{2}c$. Then I could take the $L$-th power to get
$\displaystyle \left (\frac{n(n-1)}{2}c \right )^L = \Big( 1 + (1+c) + \ldots + (1 + (N-1)c)\Big )^L = \sum {L \choose{l_1 , \ldots , l_N}} 1^{l_1}(1+c)^{l_2} (1+2c)^{l_3} \cdot \ldots \cdot  (1+(N-1)c)^{l_{N}}$
where the last line uses the multinomial formula. Then I can compute the right-hand-side by computing the much easier left-hand-side. Except the right-hand-side is the wrong series! 
Is there any trick to get rid of these pesky coefficients? One idea I had was to replace each $(1+mc)$ with some function which, when differentiated $l_m$ times adds an exponent of $l_m$ and a constant factor that cancels out the multinomial coefficients in the product rule. That would work if the left-hand-side looked like some easy to differentiate function. Unfortunately I already know what that function looks like, and it is the function that I derived the series from.
Edit: If we write $[L,N]$ for the sum above he satisfies the recurrence $[L,N] = (1+(N-1))[L-1,N]$ + [L,N-1] if anyone has ideas for that. Generating functions maybe?
 A: For $c=1$ we have
$\displaystyle \sum_{} 1^{l_1}2^{l_2} 3^{l_3} \cdot \ldots \cdot  N^{l_{N}} = {{N+L} \brace N}$ 
are the Stirling numbers of the second kind.
Write $(L,N)$ for the sum above. For each $(l_1, l_2 \ldots , l_N)$ we have either $l_N \ge 1$ or $l_N = 0$. The sequences with the first property correspond to the sequences $(l_1,l_2, \ldots , l_{N-1}, l_N-1)$ with sum $L-1$. It follows the sum over all such sequences is $N(L-1,N)$. The sequences with the second property correspond to the sequences $(l_1,l_2,\ldots, l_{N-1})$ with sum $L$. It follows the sum over all such sequences is $(L,N-1)$. Thus we have the recurrence $(L,N) = $ $N(L-1,N) + (L,N-1)$.
By iterating the recurrence we can express any given $(L,N)$ in terms of some collection of values $(1, \widetilde N)$ and $(\widetilde L,1)$. It follows any doubly-indexed sequence $C^L_N$ that satisfies the same recurrence and has all $C^L_1 = (L,1)$ and $C^1_N = (1,N)$ must have all $C^L_N = (L,N)$.
For a similar reason we see $C^L_N =  {{N+L} \brace {N}}$ satisfies the recurrence. It remains to show each $(L,1) = {{L+1} \brace 1} $ and $(1,N) = {{N+1} \brace N }$.
Clearly each $(L,1) = 1$. Since there is exactly one way to partition a set of $L+1$ elements into exactly one subset we have ${L+1 \brace 1}=1$ as well.
By definition each $(1,N) = 1 + 2 + \ldots + N = N(N+1)/2$ by the well-known formula. To compute ${{N+1} \brace N }$ observe each partition of $N+1$ into $N$ subsets consists of $N-1$ singletons and $1$ doubleton. Thus ${{N+1} \brace N }$ is just the number of doubletons of $N+1$ which is just ${N+1 \choose 2} = N(N+1)/2$. This completes the proof.
