# Similarity between $e^x$ power series and Gamma function integral?

The power series for $$e^x$$ is as follows.

$$e^{x} =\sum ^{\infty }_{n=0}\frac{x^{n}}{n!}$$

If we define $$n! = \Gamma(n+1)$$, then we have

$$n!=\int ^{\infty }_{0} x^{n} e^{-x} dx.$$

An extremely surface level observation is that if we compare the summand/integrand of each equation to their corresponding left-hand-sides, we end up with the following correspondences.

$$e^x \longleftrightarrow \frac{x^n}{n!} \hspace{2 in} n! \longleftrightarrow \frac{x^n}{e^x}$$

Very naïvely, one could make the observation that the correspondence on the left looks like an algebraic manipulation of the correspondence on the right (multiply both sides by $$n!$$ and divide by $$e^x$$).

Is this merely coincidence or is there some deeper connection here?

The exponential (actually $$e^{-x}$$ for convergence reasons) and the $$\Gamma$$ are Mellin pairs so one transforms into the other with the Mellin and the Mellin inverse transforms.
The second equation above for arbitrary $$s-1,\hspace{.1 in} \Re(s) > 0$$ rather than $$n$$ is the definition of the Mellin transform, so it shows that "$$(s-1)!" = \Gamma(s)=\int_{0}^{\infty}x^{s-1}e^{-x}dx, \hspace{.1 in} \Re(s) > 0$$ is the direct Mellin transform of $$e^{-x}$$ and one of the reasons for the shift in definition ($$n! = \Gamma(n+1)$$).
For the inverse it exactly follows by showing the coefficients of the resultant Taylor series are the coefficient of the exponential ones which is analytically reflected in $$e^{-x}=\frac{1}{2\pi i}\int_{c-\infty}^{c+\infty}x^{-s}\Gamma(s)ds, \hspace{.1 in} x>0, \hspace{.1 in}c>0$$ arbitrary.