Expand $$\frac{e^{2z}}{(z-1)^3}$$ around $z=1$
$$\frac{e^{2z}}{(z-1)^3}=\frac{e^{2z-2+2}}{(z-1)^3}=\frac{e^{2(z-1)+2}}{(z-1)^3}=\frac{e^{2(z-1)}e^2}{(z-1)^3}$$
Now can we say that $e^{2(z-1)}=(e^{(z-1)})^{2}=\sum_{n=0}^{\infty} (\frac{(z-1)^n}{n!})^2=\sum_{n=0}^{\infty} (\frac{(z-1)^{2n}}{(n!)^2})?$
$$ e^{2(z-1)}=e^{(z-1)^{2}} $$ is false! But you don't need that equation. Use the power series of the exponential function $$ e^x = \sum_{k=0}^\infty\frac1{k!}x^k\hspace{2cm}(*1) $$ to conclude $$ e^{2(z-1)}=\sum_{k=0}^\infty\frac1{k!}(2(z-1))^k = \sum_{k=0}^\infty \frac{2^k}{k!}(z-1)^k. $$
To "restrictions":
The formula $(*1)$ is given for all $x\in\mathbb C$. Hence, you can replace it by any formula and it still holds. You can even write $$ e^{\log(z-1)}=\sum_{k=0}^\infty\frac1{k!}(\log(z-1))^k. $$ Naturally, you have to restrict $z$ because of the logarithm but not because of the exponential function!
On the other hand, $$ \log(1+x)=\sum_{k=1}^\infty\frac{(-1)^k}kx^k\hspace{2cm}(*2) $$ holds just for $|x|<1$. Therefore, if you like to replace $x$ then you have the further restriction $|x|<1$. Although $z-1$ is defined for all $z\in\mathbb C$, if you like to replace $x=z-1$, you will need the further restriction $1>|x|=|z-1|$. This gives you $$ \log(z)=\log(1+z-1)=\sum_{k=0}^\infty\frac{(-1)^k}k(z-1)^k $$ for $|z-1|<1$.
(Ok, precisely the power series $(*2)$ is well defined even for some points on the boundary $|x|=1$. But it is definitely wrong for $|x|>1$. I want to point out where I thought of restrictions and simplified the details)