<s>[INST] Deriving $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ When I first saw that $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ I wondered how does one find/derive something like that (not just prove it by induction), now that I have some very basic knowledge about polynomials I gave it a try.   
Im asking if there are mistakes in the following. Thanks in advance to anyone who will bother checking that.

Let $n$ be positive integer, then:
$x^n-1$ is polynomial of degree $n$ where the coefficients of the terms between $x^n$ and $-1$ (the constant term) are equal to $0$.  
$$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$$ 
where $a_n=1, a_0=-1$ $\,$ and $\,$ $a_i=0$ for every   $i=n-1,...,1$
If $n$ is even we have two roots, $1$ and $-1$
If $n$ is odd, $1$ is the only root.   
In both cases we can rewrite it as   
$$f(x)=x^n-1=(x-1)(b_{n-1}x^{n-1}+...+b_1x+b_0)=\\ x(b_{n-1}x^{n-1}+...+b_0)-(b_{n-1}x^{n-1}+...+b_0)=\\ (b_{n-1}x^n+b_{n-2}x^{n-1}+...+b_1x^2+b_0x)-b_{n-1}x^{n-1}-b_{n-2}x^{n-2}...-b_1x-b_0\tag{A}$$
After rearranging we get:
$$f(x)=b_{n-1}x^n+(b_{n-2}-b_{n-1})x^{n-1}+(b_{n-3}-b_{n-2})x^{n-2}+...+(b_0-b_1)x+(-b_0)$$
The highest term, namely $a_nx^n$ comes from multiplying $x$ with $b_{n-1}x^{n-1}$, that is 
$$a_nx^n=x(b_{n-1}x^{n-1})=b_{n-1}x^n$$
We can conclude that $a_n=b_{n-1}=1$   
$f(0)=-1=(-1)(0+...+0+b_0)$, hence $b_0=1$.
Then the constant term $(-b_0)=a_0=-1$
Then the coefficients $(b_{i-1}-b_{i})=a_{i}=0$ for every $i=n-1,...,1$
We observe the following, $\, b_{n-2}-b_{n-1}=b_{n-2}-1=0,\,$ then $\, b_{n-2}=1$
the next coefficient,  $b_{n-3}-b_{n-2}=b_{n-3}-1=0,\,$ then  $\,b_{n-3}=1$
We can keep going like that with the other coefficients. From that we conclude that $b_i=1$ for every $i=n-1,...,1$   
Hence  $$f(x)=x^n-1=(x-1)(x^{n-1}+...+x+1)$$

At $(A)$ I use the Polynomial remainder theorem to get $(x-1)$ as factor.
I wasnt able to find something like that for $x^n+1$ when $n$ is even, there are no real roots.
Prabably could be writen like that:
 $x^n+c=(x^n-1)+c+1\;$ where $c$ is some integer.
 [/INST] You are making it much too complicated.
By multiplication of the polynomials,
$$(x-1)(x^{n-1}+x^{n-2}+x^{n-3}+\cdots x^2+x+1)=\\
x^n+x^{n-1}+x^{n-2}+\cdots\ \ \ \ x^3+x^2+x\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -x^{n-1}-x^{n-2}-x^{n-3}\cdots-x^2-x-1=\\x^n-1.$$
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