That's the correct series. You can also get it directly from the formula:
$$ f(z) = \sum_{n = 0}^\infty \frac{f^{(n)}(1)}{n!}(z - 1)^n. $$
Both methods are good.
For the second question, you have
$$ e^{1} \sum_{n=0}^{\infty} \frac{|(z - 1)^n|}{n!} \leq e \sum_{n=0}^{\infty} \frac{R^n}{n!} $$
and the ratio test shows that this converges. Thus the Taylor series converges absolutely for every $|z - 1| < R$, for every $R > 0$. Thus it converges for every $z \in \mathbb C$ (since you can always find an $R$ such that $|z - 1| < R$). This is perfectly valid reasoning.
A common argument here is exactly what you did, but phrased using the Weierstrass M-test. Namely, take
$$ f_n(z) = \frac{e^1}{n!}(z - 1)^n \text{ and } M_n = \frac{e^1}{n!} R^n $$
Then on the set $\{|z - 1| \le R\}$,
$$ |f_n(z)| \le M_n $$
Since the sum $\sum_{n \ge 0} M_n$ converges, the M-test says that
$$ \sum_{n = 0}^\infty f_n(z) = \sum_{n = 0}^\infty \frac{e^1}{n!}(z - 1)^n $$
converges absolutely and uniformly on the same set, $\{|z - 1| \le R\}$. Absolute convergence you've seen, uniform convergence implies, for instance, that the resulting sum is continuous everywhere.
Since $e^z = e^{z_0} \cdot e^{z-z_0}$, we have
$$
e^z = e^{z_0} \cdot \sum_{k=0}^{\infty} \frac{(z-z_0)^k}{k!} =
e^{1+i}\cdot\sum_{k=0}^{\infty} \frac{(z-1-i)^k}{k!}.
$$
Best Answer
You are on the right track. Since $\sinh(z-i\pi)=-\sinh(z)$ we can write
$$\sinh(z)=-\sinh(\color{blue}{z-i\pi})=-\sum_{n=0}^\infty \frac{(\color{blue}{z-i\pi})^{2n+1}}{(2n+1)!}$$