Power series that satisfies $f^{(k)}(0) = (k!)^2$ To solve a larger problem I must solve the sub-problem of finding a power series that satisfies: 
$$f^{(k)}(0) = (k!)^2.$$
My textbook states that this is "obviously"
$$\sum^{\infty}_{k=0}k!z^k.$$
Why is this? 
 A: For any power-series $\sum_{k=0}^{\infty}a_k x^k$ one has that $f^{(k)}(0)= k! a_k$.  
This is true in a 'formal' sense for any sequence of coefficients, and it is true in the analytic sense of derivative if the power-series has positive radius of convergence. 
A: Hint 
$$f(z)=\sum_{n=0}^{\infty} a_n z^n\implies a_n= \frac{f^{(n)}(0)}{n!} $$
A: Disclaimer: This does not answer the posed question (which has already been answered) but it does explain how to convert your non-convergent power series into a function satisfying your differential equations.

Your series doesn't converge for any $z\ne0$. Instead, consider the Borel sum of your series:
$$\mathcal B\sum_{n=0}^\infty n!z^n=\int_0^\infty\exp(-t)\left(\sum_{n=0}^\infty z^nt^n\right)~\mathrm dt=\int_0^\infty\frac{\exp(-t)}{1-zt}~\mathrm dt=\frac1z\exp\left(-\frac1z\right)\operatorname{Ei}\left(\frac1z\right)$$
when $\Re(z)<0$ and where $\operatorname{Ei}(x)$ is the principle exponential integral.
$$\operatorname{Ei}(x)=P.V.\int_{-x}^\infty-\frac{e^{-t}}t~\mathrm dt$$
This function happens to satisfy
$$\lim_{z\to0}\frac{\mathrm d^n}{\mathrm dz^n}\frac1z\exp\left(-\frac1z\right)\operatorname{Ei}\left(\frac1z\right)=(n!)^2$$
As intended.
