Does $\big(\cos\frac{\pi}4+i\sin\frac{\pi}4\big)^n = \cos\frac{n\pi}4+i\sin\frac{n\pi}4$ for all $n$? This was the original question:

Prove that $(1+i)^{10}=32i$ for $i=\sqrt{-1}$ using only trigonometry.

It turned out to be an interesting problem, which then led to another question instead.
Firstly, from Euler's identity, I know that $i=e^{\pi i\div 2}$ $$\therefore 1+i=\sqrt 2\bigg(\cos\frac{\pi}4 + i\sin\frac{\pi}{4}\bigg)$$
I know $\sqrt{2}^{10}=32$ so it suffices to prove that $$i=\bigg(\cos\frac{\pi}4 + i\sin\frac{\pi}{4}\bigg)^{10}$$
And this was where I got stuck.
However, working backwards from $32i$, it led me to a conjecture. $$32i=32(0+i)=32\bigg(\cos\frac{\pi}2+i\sin\frac{\pi}{2}\bigg)$$ and now in order to make the denominators equal to $4$ without changing the value of the brackets, I decided to do it like this. $$\cos\frac{\pi}2+i\sin\frac{\pi}{2}=\cos\bigg(2\pi+\frac{\pi}2\bigg)+i\sin\bigg(2\pi+\frac{\pi}{2}\bigg)=\cos\frac{10\pi}4+i\sin\frac{10\pi}{4}$$ $$\therefore \cos\frac{10\pi}4+i\sin\frac{10\pi}{4}=\bigg(\cos\frac{\pi}4+i\sin\frac{\pi}4\bigg)^{10}$$
Can the last equation be generalised for some $n$, i.e.

$$\bigg(\cos\frac{\pi}4+i\sin\frac{\pi}4\bigg)^n \stackrel{\small ?}{=}\cos\frac{n\pi}4+i\sin\frac{n\pi}4$$

Or is this just a coincidence for $n=10$? Can the denominator also be changed arbitrarily and not just be fixed on $4$?
How does one go about proving this using only trigonometry? I can't apply the binomial theorem because then I need to use the gamma function, which I don't believe falls under trigonometry in the sense of the question (it was a challenge set by a teacher at school actually, but just for fun and not homework-based). So what other methods are there?
Also, I asked my teacher where he found this problem and he wouldn't tell me lest I would search up the answer, and... well, it seems I am... ahem. So hints would be much appreciated.
Any ideas? Thanks.
 A: This follows easily from Euler's formula. 
$\begin{align*}
\left(\cos(\theta)+i\sin(\theta)\right)^n&=\left(e^{i\theta}\right)^n=e^{ni\theta}=\cos(n\theta)+i\sin(n\theta)
\end{align*}$
Although the above looks almost trivial, it has a cool interpretation for trigonometry. We have $\cos(\theta)=\mathrm{Re}(e^{i\theta})$ and $\sin(\theta)=\mathrm{Im}(e^{i\theta})$. So if we wanted to deduce the double angle formula for example, we can simply plug in $2\theta$,
$\begin{align*}\cos(2\theta)=\mathrm{Re}((e^{i\theta})^2)&=\mathrm{Re}\big[\big(\cos(\theta)+i\sin(\theta)\big)^2\big]\\&=\mathrm{Re}\big(\cos^2(\theta)+2i\cos(\theta)\sin(\theta)-\sin^2(\theta)\big)\\&=\cos^2(\theta)-\sin^2(\theta)\end{align*}$
$\sin(2\theta)=\mathrm{Im}\big(\cos^2(\theta)+2i\cos(\theta)\sin(\theta)-\sin^2(\theta)\big)=2\cos(\theta)\sin(\theta)$ 
I bet they didn't show you that proof in trigonometry class! :D
In general, this gives us a formula for $\cos(n\theta)$ and $\sin(n\theta)$ which has an obvious geometric intepretation. The formula just follows from the binomial theorem. However, it's easy not to recognize because the factor of $i^n$ ends up separating the even and odd terms into the identities for cosine and sine, respectively. 
