Rearranging exp terms
Hint:

So you have proved that
$$\exp\left({j\frac \pi4 n}\right)+ \exp\left({j\frac{7\pi} 4 n}\right) =\exp\left({-j\frac \pi4 n}\right)+ \exp\left({j\frac \pi4 n}\right)$$

The second derivation does results in the same answer as well. It is not prohibited but one would want to simplify things here. It does not help factorizing with a factor $\exp({j\pi n})$ because $\exp({j\pi n}) = \left(-1\right)^n$ and not $-1$. But one could factor by $\exp({j2\pi n})$ for instance since $\exp({j2\pi n}) = 1$. With this in mind:


$$\exp\left({j\frac\pi4n}\right)+ \exp\left({j\frac{7\pi}{4}n}\right)={\exp\left({j2\pi n}\right)}\cdot\left[\underbrace{\exp\left(-j\frac{7\pi}{4}n\right)}_{\exp\left(j\frac{\pi}{4}n\right)}+ \exp\left(-j\frac{1\pi}{4}n\right)\right]$$
And the two forms are identical.