Rearranging exp terms I have some problems rearranging exp terms to receive a cos or sin. I guess the solution is silly but I‘m not able to spot my mistake. 

I know the solution (since i looked it up with wolfram alpha) has to be a $cos(\pi /4n)$ and therefore  $\frac{e^{-j\pi/4n}+ e^{j\pi/4n}}{2}$.
My goal is to rearrange the two terms $e^{j\pi/4n}+ e^{j7/4\pi n}$ (devided by 2) to receive this cos. 
One can clearly see that by removing the $\pi/4$ e term on both sites of 
$$e^{j\pi/4n}+ e^{j7/4\pi n} =e^{-j\pi/4n}+ e^{j\pi/4n}$$
this remains
$$ e^{j7/4\pi n}= e^{-j\pi/4n}$$ wich is true, when its drawn on the unit circle.

If one doesnt see this trick, my usual approach is to split an e term up to make the remaining two e terms conj. compex of each other. Like this
$$ e^{j\pi n} \cdot (e^{-j3\pi/4n}+ e^{j3\pi/4n})$$

Saddly this results (as far as I can see) in $(-1)^n\cdot 2cos(3/4\pi n)$, which is not what I‘ve expected. 
How comes this rearrangement doesn‘t result in the same cos? Is this rearrangement in some cases prohibited?
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.