I have the following function:
$$f(x)=\left(T_{N_2}(x)-T_{N_1}(x)\right)\left(T_{N_3}(x)-T_{N_1}(x)\right)\left(T_{N_3}(x)-T_{N_2}(x)\right)$$
where $T_{N}(x)=1+x+\frac{x^2}{2!}+...+\frac{x^N}{N!}$
Looking at the difference yields
$$T_{N_2}(x)-T_{N_1}(x)=\sum_{k=N_1+1}^{N_2}\frac{x^k}{k!}$$
Therefore my polynomial becomes
$$f(x)=\sum_{k=N_1+1}^{N_2}\frac{x^k}{k!}\sum_{k=N_1+1}^{N_3}\frac{x^k}{k!}\sum_{k=N_2+1}^{N_3}\frac{x^k}{k!}$$
I would like to simplify further from here, but i'm running into a problem. For example, when $N_1=1, N_2=4,$ and $N_3=6$ the polynomial is
$$\frac{x^9}{2!2!5!}+\left(\frac{2}{2!3!5!}+\frac{1}{2!2!6!}\right)x^{10}+\left(\frac{2}{2!3!6!}+\frac{2}{2!4!5!}+\frac{1}{3!3!5!}\right)x^{11}+\left(\frac{2}{2!4!6!}+\frac{1}{2!5!5!}+\frac{2}{3!4!5!}+\frac{1}{3!3!6!}\right)x^{12}+\left(\frac{2}{2!5!6!}+\frac{1}{3!5!5!}+\frac{2}{3!4!6!}+\frac{1}{4!4!5!}\right)x^{13}+\left(\frac{1}{2!6!6!}+\frac{2}{3!5!6!}+\frac{1}{4!4!6!}+\frac{1}{4!5!5!}\right)x^{14}+\left(\frac{2}{4!5!6!}+\frac{1}{3!6!6!}\right)x^{15}+\frac{x^{16}}{4!6!6!}$$
There is a combinatorial argument, i think, to be had with the different denominators of the polynomial's coefficients, but I'm not sure what it is. Obviously the lowest degree coefficient's denominator is just $(N_1+1)!(N_1+1)!(N_2+1)!$ and a similar argument for the leading coefficient as well with $N_2!N_3!N_3!$
Is there any way to come up with a summation formula (could be a double or a triple) for this combinatorial argument?
Taking your polynomial as Exponential generating function, the coefficient of $\frac{x^n}{n!}$ in $$ \left( \sum_{i=N_1 + 1}^{N_2}\frac{x^i}{i!} \right) \left( \sum_{i=N_1 + 1}^{N_3}\frac{x^i}{i!} \right) \left( \sum_{i=N_2 + 1}^{N_3}\frac{x^i}{i!} \right) $$ is the number of ways to split a group of $n$ people into three groups, the first of size between $N_1+1$ and $N_2$, the second with size between $N_2 + 1$ and $N_3$, the third with size between $N_1$ and $N_3$. You can write this as, e.g. $$ \text{coeff of } \frac{x^n}{n!} = \sum_{i=N_1 + 1}^{N_2} {n \choose i} \sum_{j = \max(N_1 + 1, n - i - N_3)}^{\min(N_3, n - i - N_2 - 1)} {n - i \choose j} $$ or with trinomial coefficients your function is $$ \sum_{i = N_1 + 1}^{N_2} \sum_{j = N_1 + 1}^{N_3} \sum_{k = N_2 + 1}^{N_3} {i + j + k \choose i, j, k} \frac{x^{i+j+k}}{(i+j+k)!} $$ But it's really not going to get any cleaner than that.