Taylor series of $e^{(x-1)^2}$ about $x=1$

Question

How would we find the Taylor series of $e^{(x-1)^2}$ about $a=1$?

I tried finding the answer using the Taylor series of $e^x$ about $a=1$ which I was able to do correctly. When I substituted $(x-1)^2$ back in for $x$, things went wrong.

Answer 1

$$e^{x}=\sum_{n=0}^\infty {x^n\over n!}\implies e^{\color{blue}{(x-1)^2}}=\sum_{n=0}^\infty {(\color{blue}{(x-1)^2})^n\over n!}=\sum_{n=0}^\infty {(x-1)^{2n}\over n!}$$