Although there is already an answer of mine, I want to add another answer.

This is **TL;DR**.

*Short answer.* This is because logarithmic function lacks factorial in its Taylor expansion.

*Medium answer.* Riemann's functional equation links exponential and trigonometric functions with logarithms and inverse trigonometric. It contains everything what you need to make an exponent from a logarithm.

*Long answer.*

This is Taylor series for logarithm:

$$\ln(z+1)=z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+\frac{z^5}{5}-\frac{z^6}{6}+\frac{z^7}{7}-\frac{z^8}{8}+\frac{z^9}{9}-\frac{z^{10}}{10}+O\left(z^{11}\right)$$

This is Taylor series for exponent:

$$\exp (z)-1=z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\frac{z^5}{5!}+\frac{z^6}{6!}+\frac{z^7}{7!}+\frac{z^8}{8!}+\frac{z^9}{9!}+\frac{z^{10}}{10!}+O\left(z^{11}\right)$$

What should we add to the former to get the later? Well, we have to add the factorial and remove the counter from the denominator.

Consider such algebraic element $\omega_+$ (not a real number) on which a function "standard part" is implemented in such a way, that $\operatorname{st} \omega_+^n=B_n^*$ where $B_n^*$ are Bernoulli numbers (with $B_1^*=1/2$), or more generally, $\operatorname{st}\omega_+^x=-x\zeta(1-x)$.

Now consider the function

$$\frac{z}{2\pi} \log \left(\frac{\omega _+-\frac{z}{2 \pi }}{\omega _++\frac{z}{2 \pi }}\right)$$
Its Taylor series is

$$-\frac{z^2}{2 \left(\pi ^2 \omega _+\right)}-\frac{z^4}{24 \left(\pi ^4 \omega _+^3\right)}-\frac{z^6}{160 \left(\pi ^6 \omega _+^5\right)}-\frac{z^8}{896 \left(\pi ^8 \omega _+^7\right)}-\frac{z^{10}}{4608 \left(\pi ^{10} \omega _+^9\right)}+O\left(z^{11}\right)$$

Following Riemann's functional equation and our definition, we have:

$$\operatorname{st}\omega_+^{-x}=\operatorname{st}\frac{-\omega_+^{x+1} 2^x\pi^{x+1}}{\sin(\pi x/2)\Gamma(x)(x+1)}$$

So we can substitute the negative powers of $\omega_+$ with positive powers without changing the standard part of the whole expression.

The non-zero terms are

$$\frac{2 \left(-\frac{1}{2 \pi }\right)^n \left(-\omega _+\right){}^{1-n}}{n-1}$$

and after substitution we have

$$\frac{\omega _+^n \sec \left(\frac{\pi n}{2}\right)}{\Gamma (n+1)}$$

The resulting series is

$$\frac{1}{2} \omega _+^2 z^2+\frac{1}{24} \omega _+^4 z^4-\frac{1}{720} \omega _+^6 z^6+\frac{\omega _+^8 z^8}{40320}-\frac{\omega _+^{10} z^{10}}{3628800}+O\left(z^{11}\right)$$

oh, wait... is not it similar to

$$\cos \left(\omega _+ z\right)=1-\frac{1}{2} \omega _+^2 z^2+\frac{1}{24} \omega _+^4 z^4-\frac{1}{720} \omega _+^6 z^6+\frac{\omega _+^8 z^8}{40320}-\frac{\omega _+^{10} z^{10}}{3628800}+O\left(z^{11}\right)$$

Well, we got:

$$\operatorname{st}\frac{z}{2 \pi } \log \left(\frac{\omega _+-\frac{z}{2 \pi }}{\omega _++\frac{z}{2 \pi }}\right)=\operatorname{st}(\cos \left(\omega _+ z\right)-1)$$

In a similar way one can establish other impressive relations:

$$\operatorname{st}(\exp \left(\omega _+ z\right)-\omega _+ z-1)=\operatorname{st}\frac{i z}{2 \pi } \log \left(\frac{\omega _+-\frac{i z}{2 \pi }}{\omega _++\frac{i z}{2 \pi }}\right)$$

$$\operatorname{st}\cos \left(\omega _+ z\right)=\operatorname{st}\frac{ z}{2 \pi } \log \left(\frac{\omega _+-\frac{ z}{2 \pi }}{\omega _-+\frac{ z}{2 \pi }}\right)$$

$$\operatorname{st}\cosh \left(\omega _+ z\right)=\operatorname{st}\frac{i z}{2 \pi } \log \left(\frac{\omega _+-\frac{i z}{2 \pi }}{\omega _-+\frac{i z}{2 \pi }}\right)$$

(where $\omega_-=\omega_+-1$).

In other words, Riemann's functional equation is a direct bridge that connects exponential function to logarithm, trigonometric functions to inverse trigonometric, transforming each term of the series separately.

conceptuallyrelated to sums of powers. The $\zeta$ function itself is defined as a non-alternating sum of powers for $\Re(z)>1$, and as an alternating sum of powers (times a certain factor) for $\Re(x)\in(0,1)$ On the other hand, geometric shapes of the form $x^n+y^m=1$, calledsuperellipsesorLame curves, are also bounded sums of powers. But by integrating $y=\sqrt[m]{1-x^n}$ or $x=\sqrt[n]{1-y^m}$ on $(0,1)$ we get the multiplicative inverse of the binomial coefficient ${m+n\choose n}={m+n\choose m}$, which is obviously expressible in terms of the $\Gamma$ function. $\endgroup$ – Lucian Jun 1 '14 at 16:52