Reaching the gamma function from the Taylor series for $e^x$?

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The gamma function is expressed as $n! =\Gamma(n+1)= \int_{0}^{\infty}x^ne^{-x}dx$

which reminds me of the Taylor series for $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$

Unmathematically, this makes the gamma function make sense to me: you're effectively doing a 'rearrangment' to get $n!$ However, if I am right about my intuition, I'm feeling particularly foolish about it, as I can't for the life of me properly explain what is precisely going on to get you from one to the other.

So, to you more experienced mathematicians out there, am I correct about my intuition, and if so, could I possibly trouble you to tell me exactly how you get from the Taylor series to the gamma function?

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It is true that on one hand,

$$\int_0^\infty\frac{x^n}{n!}e^{-x}dx=1$$ and on the other

$$\sum_{n=0}^\infty\frac{x^n}{n!}e^{-x}=1.$$

But the integral is on $x$, while the sum is on $n$, and I just see a coincidence here.


You can modify the argument of the exponential to ensure convergence and establish for $s>1$

$$\sum_{n=0}^\infty\int_0^\infty\frac{x^n}{n!}e^{-sx}dx=\sum_{n=0}^\infty s^{1-n}=\frac1{s-1}=\int_0^\infty e^{(1-s)x}dx=\int_0^\infty\sum_{n=0}^\infty\frac{x^n}{n!}e^{-sx}dx.$$