I'm not yet very good at complex number, so I would appreciate the following insight:
How exactly do we arrive from $e^{\pi(1-i)}-e^{-\pi(1-i)}$ to $e^{-π}-e^π$, and why does $e^{\pi(1-i)n}-e^{-\pi(1-i)n}$ have different signs for $n$ odd and $n$ even?
On
Recall that, for any $w, z \in \Bbb C$,
$e^{w + z} = e^w e^z \tag{1}$
and of course the Euler formula
$e^{i\theta} = \cos \theta + i \sin \theta; \tag{2}$
then via (1),
$e^{\pi(1-i)}-e^{-\pi(1-i)} = e^{\pi - i\pi} - e^{i\pi - \pi} = e^\pi e^{-i\pi} - e^{-\pi}e^{i\pi} = e^{-\pi} - e^\pi, \tag{3}$
since, by (2),
$e^{i\pi} = \cos \pi + i \sin \pi = -1; \tag{4}$
likewise,
$e^{\pi(1-i)n}-e^{-\pi(1-i)n} = e^{n\pi} e^{-i \pi n} - e^{-n\pi} e^{i\pi n}; \tag{5}$
but
$e^{i\pi n} = e^{-i\pi n} = \cos n \pi = 1 \; \text{for even} \;n \; \text{or} \; -1, \; \text{when} \; n \; \text{is odd}. \tag{5}$
For your first question, you must know by now that $e^{i\pi}=-1$, if you do not, then note that it follows from Euler's identity, since $$ e^{i\pi}=\cos(\pi)+i\sin(\pi) = -1+i0=-1 \tag{1} $$ Hence \begin{align} e^{\pi(1-i)}-e^{-\pi(1-i)} & = e^{\pi}e^{-\pi i}-e^{-\pi}e^{\pi i} \\ & = \frac{e^{\pi}}{e^{\pi i}}-e^{-\pi}e^{\pi i} \\ & \overset{(1)}{=} -e^\pi+e^{-\pi} \\ &= e^{-\pi} -e^\pi \end{align}
For the second one, a similar line of thought shows that $$ e^{\pi(1-i)n}-e^{-\pi(1-i)n} = e^{\pi n}(-1)^n - e^{-\pi n}(-1)^n $$ Thus indeed the sign varies depending if $n$ is odd or even.