The following is a rigorous proof of De Moivre's theorem by means of mathematical induction.
The theorem put simply is that:
Any complex number, $z = a+bi$, on a cartesian plane can be expressed in polar form, where $a=r\cos\theta$ and $b=r\sin\theta$ and $r$ is the absolute distance from the origin to the point $z$. In light of this, the expression for the complex number $z$ in polar form is $z=r(\cos\theta+i\sin\theta)$
In order to find the $n^{th}$ power of $z$, the following rule applies:
$$\displaystyle z^n=(r(\cos\theta+i\sin\theta))^n=r^n(\cos n\theta+i\sin n\theta)$$
where $n\in\mathbb N$ (I have confined $n$ to the natural numbers given that I intend on using mathematical induction to prove the theorem.
Base step:
We must prove that what is stated is true for $n=1$, hence
$$z^1=r^1(\cos 1\cdot\theta+i\sin1\cdot \theta)$$
$$=r(\cos\theta+i\sin\theta)$$
which is true. Next we assume it is true for $n=k$, and therefore proceed to the inductive step.
Inductive step:
When calculating $z^{n+1}$, it's the same as calculating $z^nz$, hence
$$z^{n+1}=(r^n(\cos n\theta + i\sin n\theta))(r(\cos \theta+i\sin \theta))$$
$$=r^nr(\cos n\theta+i\sin n\theta)(\cos\theta +i\sin \theta)$$
$$=r^{n+1}(\cos n\theta\cdot\cos\theta+\cos n\theta \cdot i\sin\theta+i\sin n\theta\cdot\cos\theta+i^2\sin n\theta\cdot\sin\theta)$$
from here, $i^2=-1$, and therefore $+i^2\sin n\theta\cdot\sin\theta$ becomes $-\sin n\theta\cdot\sin\theta$. In addition to this, the following trigonometric identities will be used:
$$\sin(n\theta + \theta) = \sin n\theta\cdot\cos\theta+\cos n\theta\cdot\sin\theta$$
$$or$$
$$k\sin(n \theta+\theta)=k\sin n \theta\cdot\cos \theta + \cos n\theta\cdot k\sin\theta$$
$$and$$
$$\cos(n \theta + \theta)=\cos n\theta\cdot\cos\theta-\sin n\theta\cdot\sin\theta$$
by using these, the original equation has now become
$$=r^{n+1}(\cos(n \theta + \theta) + i\sin(n \theta + \theta))$$
then, after factoring out $\theta$, the theorem becomes
$$=r^{n+1}(\cos((n+1)\theta)+i\sin((n+1)\theta))$$
which completes the proof.
Question: is this proof correct all the way through? Or have I missed any conditions? If all is well in the case of this proof, could somebody aid me in proving it for $\{n|n\in\mathbb R,n\geq1\}$
You're very close!
The first issue I see is a minor one. The identity $$k\sin(n \theta+\theta)=k\sin n \theta\cdot\cos \theta + \cos n\theta\cdot k\sin\theta$$ seems to be unnecessary. I just can't tell what use you make of it, since the preceding identity does what you need.
The second issue I see is a major one. The identity $$\cos(n \theta + \theta)=\cos n\theta\cdot\cos\theta+\sin n\theta\cdot\sin\theta$$ is incorrect. It should instead be $$\cos(n \theta + \theta)=\cos n\theta\cdot\cos\theta-\sin n\theta\cdot\sin\theta,$$ which allows you to draw the desired conclusion. (I suspect this may have been a typo.)