derivation of discrete Fourier Transform I'm learning DSP on my own. Found a free online book. Currently reading a chapter on Discrete Fourier Transform. Right out of the gate I can't understand several things. I'm pretty good with math and have a decent understanding of complex numbers and their representation with the help of $e^{j\phi}$. But this got me stumbled.

Below is the paragraph I can't understand and the questions pertaining to it.


  We now develop a Fourier representation for finite-length signals; to
  do so, we need to find a set of oscillatory signals of length $N$
  which contain a whole number of periods over their support. We start
  by considering a family of finite-length sinusoidal signals (indexed
  by an integer $k$) of the form
  
  $$ w_k[n]=e^{j \omega_k n}, \space n = 0,...,N-1 $$ where all the $\omega_k$’s are
  distinct frequencies which fulfill our requirements. To determine
  these frequency values, note that, in order for $w_k[n]$ to contain a
  whole number of periods over $N$ samples, it must conform to
  
  $$ w_k[N]=w_k[0]=1 $$ which translates to $$ (e^{j \omega_k})^N = 1 $$
  
  The above equation has $N$ distinct solutions which are the $N$ roots
  of unity $e^{j\frac{2\pi m}{N}}$, $m = 0,…,N - 1$ 



which contain a whole number of periods over their support.
What does "periods of their support" mean?
note that, in order for $w_k[n]$ to contain a whole number of periods over $N$ samples. Why does $w_k[n]$ must contain a whole number of periods over $N$ samples? Where does this requirement come from?
it must conform to $w_k[N]=w_k[0]=1$. Why? Where does this requirement come from? I do understand $w_k[N]=w_k[0]$, that's because it is periodic, what I don't understand is $=1$ portion. Why at sample $0$ and sample $N$ it must be equal to 1?
which translates to $(e^{j \omega_k})^N = 1$. How does $w_k[N]=w_k[0]=1$ translate to $(e^{j \omega_k})^N = 1$?
The above equation has $N$ distinct solutions which are the $N$ roots of unity $e^{j\frac{2\pi m}{N}}$, $m = 0,…,N - 1$. Can someone demonstrate the finding of these roots?
The Discrete Fourier Transform is one-and-the-same as the Discrete Fourier Series.  The DFT (and explicitly DFS) maps one periodic sequence, $x[n]$ of period $N$:

$$ x[n+N] = x[n] \qquad \forall n \in \mathbb{Z} $$

to another periodic sequence, $X[k]$, having the same period $N$:

$$ X[k+N] = X[k] \qquad \forall k \in \mathbb{Z} $$

and the inverse DFT (or "iDFT") maps it back.

I have a pretty rigid Nazi-like position about it (see the other answer), but I don't think that the math cares.  Now if you wanna connect this to the more common (but often misinterpreted) derivation, you can look at a previous answer of mine.

The term "support of $x[n]$" simply means the domain of values of $n$ in which $x[n]$ (presumed to be non-periodic and to have "finite support") is potentially non-zero.  It could also mean the duration of $x[n]$.  But when you periodically extend $x[n]$, then that sequence has no finite support.  $x[n]$ is defined and potentially non-zero everywhere (or all integers).