I have the following expression:
$$\sum_{i=0}^{n}\binom{n}{i}(2x+1)^{n-i}(-1)^ii!$$
Without the $i!$, the above expression would simply reduce to $(2x)^n$, but is there a way, or method for simplifying the expression which includes the extra $i!$? I mean, I could simplify to $$\sum_{i=0}^{n}\frac{n!}{(n-i)!}(2x+1)^{n-i}(-1)^i$$ But this is not really simplifying as I am trying to find a way to remove the sigma.
By setting $j=n-i$, $$\sum_{i=0}^{n}\frac{n!}{(n-i)!}(2x+1)^{n-i}(-1)^i = (-1)^n n!\sum_{j=0}^{n}\frac{(-1)^j (2x+1)^j}{j!},$$ hence your sum is just $(-1)^n n!$ times a partial sum for the Taylor series of $e^{-(2x+1)}$ in a neighbourhood of the origin.