Why does $(\cos \theta + i \sin \theta)^n =(\cos n\theta + i \sin n \theta)$ Is it the Euler identity $$ e^{i \theta} =(\cos \theta + i \sin \theta)$$
$$ e^{i n \theta} =(\cos n \theta + i \sin n \theta)$$
 A: Hints (sketches)
First Proof: Trigonometric identities + induction: $\;\;\;\;$   For $\,n=2\;$ :
$$(\cos +i\sin t)^2=\cos^2t-\sin^2t+2i\cos t\sin t=\cos 2t+i\sin 2t$$
Induction:
$$(\cos t+i\sin t)^{n+1}=(\cos +i\sin t)^n(\cos +i\sin t)\stackrel{\text{ind. Hyp.}}=(\cos nt+i\sin nt)(\cos +i\sin t)=$$
$$=\cos nt\cos t-\sin nt\sin t+i(\sin nt\cos t+\sin t\cos nt)=\ldots\ldots$$
Second "Proof": Using polar representation 
$$(\cos t+i\sin t)^n=\left(e^{it}\right)^n=e^{int}=\ldots\ldots$$
You can see that the second proof is way easier and direct than the first one...yet it requires to know some stuff that makes it so.
Choose yours...:)
A: You can prove geometrically, that if you take a right-angle triangle a:b:x, and a second triangle c:d:y, that the 'product' of these produces a ray $xy$ at the sum of the angles.
The proof runs like this.
                BD"                            Lengths are shown as 'inches' (")
             Qo---oR          OP = XC"         Angles are shown as 'degrees' (°)
                  | AD"       PQ = XD"         Points names have no extra symbol
                  |           OQ = XY"
                 Po                            OPQ is in the ratio cx:dx:yx
                  | BC"       POA = m°                ie  c:d:y.
       Oo---AC"---oA          POQ = n°  AOQ = (m+n)°

Triangle OAP and PRQ are similar triangles in the ratio of a:b:x, scaled by c and d.  
Triangle OPQ is a triangle in the ratio c:d:y, scaled by the measure x.
The line APR is straight, we can then use cartesian coordinates to define O at (0,0) etc.  Writing the right-angle coordinate as $a+bi$, it is possible to show that $i^2=-1$, when the nature of 'multiply' is taken to be add the angles, and multiply the rays.
The coordinate of OQ is $(AC-BD)+i(BC+AD)$.  When this is written as the product $(A+iB)(C+iD)$, it gives directly that $i^2=-1$.
From this, we have then that $cis(x) = \cos(x)+i \sin(x)$ has a length of one, and an angle of $x$.  The n'þ power of this has a radius of 1, and an angle of nx, ie
$$(\cos(x)+j \sin(x))^n = \cos(nx)+\sin(nx)$$
follows directly.
