Background
I think I managed to analytically continue a function $n > 1$. Can someone verify if this is correct. Let,
$$ S(0,x) = x$$ $$ S(1,x) = e^x$$ $$ S(2,x) = e^{e^x}$$ $$ S(3,x) = e^{e^{e^{x}}}$$
And so on ...
Now taking a derivative with respect to $x$ for $S(x)$ for $n\geq 1$: $$ \frac{\partial S(n,x)}{\partial x} = S(n,x)\cdot S(n-1,x)\cdot S(n-2,x)\dots\cdot S(1,x)$$
Taking a logarithm:
$$ \ln (\frac{\partial S}{\partial x}) = S(n-1,x) + S(n-2,x) + \dots + S(0,x)$$
Writing this as a complex integral and assuming the series is taylor expandable in $n$:
$$ \ln (\frac{\partial S}{\partial x}) = \oint \Big( \frac{S(z,x)}{(z-n+1)} +\frac{S(z,x)}{(z-n+2)} + \dots + \frac{S(z,x)}{z} \Big) dz $$
Factoring the $S(z,x)$ and writing as a factorial:
$$\ln (\frac{\partial S(n,x)}{\partial x}) = \oint \Big( S(z,x) \frac{\partial \ln(z! / (z-n)!)}{\partial z} dz \Big) $$
Now we know that the gamma function is the analytic continuation of the factorial function. Hence, we should be able to substitute non-integer $z$.
Questions
Is the above proof correct? (I'm not sure of the last step) I'm still not sure what, say, $S(1.5,x)$ will look like? Is there another functional equation I can use to continue this to the whole plane?