I was considering how to calculate a general solution for the following series by using the Maclaurin series;$$\sum_{n=1}^{\infty}\frac{n^a}{n!}=\sum_{n=0}^{\infty}\frac{n^a}{n!}$$ where $a{\in}{\mathbb{N}}$. Seeing the factorial in the denominator, i considered using the Maclaurin series for $e^x$ $$f(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}f^{(n)}(0).$$ So Looking at the exponential Maclaurin series, we see that $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}\;{\Rightarrow}\;\frac{d}{dx}(e^x)=\frac{d}{dx}\big(\sum_{n=0}^{\infty}\frac{x^n}{n!}\big)=\sum_{n=0}^{\infty}\frac{nx^{n-1}}{n!}.$$ Differentiation of the seriesadds a product of n into the numerator of the fraction, but decreases the x power by one, so the next product introduced to the the numerator by another itteration of differentiation would be $n-1$, which is less useful. Multiplying each side by x can help with this, so we have
$$xe^x=\sum_{n=0}^{\infty}\frac{nx^n}{n!}\;{\Rightarrow}\;\frac{d}{dx}(xe^x)=\frac{d}{dx}\big(\sum_{n=0}^{\infty}\frac{nx^n}{n!}\big)=\sum_{n=0}^{\infty}\frac{n^2x^{n-1}}{n!}$$
And this process of differentiating and multiplying by x can be repeated until the summation is in the form of $$\sum_{n=0}^{\infty}\frac{n^ax^n}{n!}$$ and can be calculated for $x=1$, which is the original summation that i aimed to calculate a result for, by finding the value of $e^x$ differentiated and then multiplied by x, repeated $a{\in}{\mathbb{N}}$ times, at x=1.
However, finding this left hand side is my problem. Let $E_0(x)=e^x$, and $E_a(x)=xE^{(r-1)}(x)$. We have that $$E_a(x)=\sum_{n=0}^{\infty}\frac{n^ax^n}{n!}\;{\Rightarrow}\;E_a(1)=\sum_{n=0}^{\infty}\frac{n^a}{n!}.$$ If I can find a formula for $E_a(1)$, then I am finished as this is the value of the summation that I am trying to calculate, but i have had no success finding a formula for $E_a(1)$.
Could someone please suggest how to find such a formula if possible?
