Complex number, entire function 
Let $f(z)=\frac{(e^{cz}-1)}{z}$ if $z\neq0$ and $f(0)=c$ show that f
  is entire

Theorem:A power series represents a analytical function inside their circle of convergence.
I know I could prove that the function is analytic without using power series, but exercise suggests it is proved that the function is analytic with power series.
How can I represent this function in power series? I know that
$$e^z=\sum_{n=0}^\infty \frac{z^n}{n!}\Rightarrow e^{cz}=\sum_{n=0}^\infty \frac{c^nz^n}{n!}\Rightarrow \frac{e^{cz}}{z}=\sum_{n=0}^\infty \frac{c^nz^{n-1}}{n!}$$
I need to separate the function?
$$f(z)=\frac{(e^{cz}-1)}{z}=\frac{e^{cz}}{z}-\frac{1}{z}$$

Entire means that is holomorphic in the whole complex plane.
Just show that the derivative of that function exists (in the complex sense of course) and you're done. Why do you care about power series?

If you want to use power series, find the power series of that function and show that it converges everywhere.
This can be done either remembering that $$e^z = \sum_{n=0}^\infty \frac {z^n}{n!}$$ and simplifying the expression or just calculating the derivatives (which is probably a little cumbersome)
Post where exactly do you have difficulties though! 
Edit
You don't need to separate the function. 
$$e^{cz}=\sum_{n=0}^\infty \frac{c^nz^n}{n!}\Rightarrow e^{cz}-1=\sum_{n=1}^\infty \frac{c^nz^{n}}{n!}$$
(notice that the index of the sum starts from $1$ now instead of $0$. You may want to convince yourself of why that is correct)
So finally $$\frac{e^{cz}-1}{z}=\sum_{n=1}^\infty \frac{c^nz^{n-1}}{n!}$$
You just need to check that in $z=0$ the series converges to $c$, but that is indeed the case.
