This question is based on the discussion in B4-1.2 of Fundamentals of Mathematics, Volume 1 Edited by H. Behnke, F. Bachmann, K. Fladt, W. Süss and H. Kunle.
See the bold text for the part of the proof I am seeking help with. My question is: how do I use the method suggested to obtain the needed result?
Technically this proof is applied to $\mathcal{R}$ which is an arbitrary commutative ring with unit element 1, and without null divisors. And the expression
$$ f\left[x\right]=\sum_{i=0}^{n}a_{i}x^{i} $$
is equivalent to an entire rational function of one argument in $\mathcal{R}.$ Here we have defined $0^{0}\equiv1$ to support our notation. For the purpose of this question we can simply call $f$ a polynomial of degree $n$ in $\mathbb{R}.$
Our objective is to prove the following theorem: If $f$ is expressible in the form
$$ f\left[x\right]=\sum_{i=0}^{n}a_{i}x^{i} $$
with $a_{n}\ne0,$ then $f$ has at most $n$ roots.
Proof: First we introduce the constant expressions
$$ \bar{a}_{k}=\sum_{i=k+1}^{n}a_{i}\alpha^{i-k-1}, $$
and the function
$$ f_{1}\left[x\right]\equiv\sum_{k=0}^{n-1}x^{k}\bar{a}_{k}=\sum_{k=0}^{n-2}x^{k}\bar{a}_{k}+a_{n}x^{n-1}. $$
The right-most expression follows from $\bar{a}_{n-1}=a_{n},$ and shows that $f_{1}$ has at most $n-1$ roots by the induction hypothesis.
Next we observe that for the real number constant $\alpha$ we have
$$ (x-\alpha)\sum_{k=0}^{i-1}x^{k}\alpha^{i-k-1}=\sum_{k=0}^{i-1}x^{k+1}\alpha^{i-k-1}-\sum_{k=0}^{i-1}x^{k}\alpha^{i-k} $$
$$ =\sum_{k=1}^{i}x^{k}\alpha^{i-k}-\sum_{k=0}^{i-1}x^{k}\alpha^{i-k}=x^{i}-\alpha^{i}, $$
and apply this result to obtain, for $n>0$
$$ f\left[x\right]-f\left[\alpha\right]=\left(x-\alpha\right)f_{1}\left[x\right]. $$
The case of $n=0$ is the constant function $f\left[x\right]=a_{0}x^{0},$ which is of the required form. If $\alpha$ is a root of $f,$ we have $$ f\left[x\right]-f\left[\alpha\right]=\left(x-\alpha\right)f_{1}\left[x\right]=0. $$
If $x\ne\alpha,$ is a root of $f$ then $f_{1}\left[x\right]=0.$ It follows that $f$ has at most one more root than $f_{1}$ Since $f_{1}$ has at most $n-1$ roots this which completes the proof.
The step I am not following is the establishment of $f\left[x\right]-f\left[\alpha\right]=\left(x-\alpha\right)f_{1}\left[x\right]$ using
$$(x-\alpha)\sum_{k=0}^{i-1}x^{k}\alpha^{i-k-1}=x^{i}-\alpha^{i}.$$ I am able to establish the result using brute force as follows: Expand $f_{1}$
$$ f_{1}\left[x\right]\equiv\sum_{k=0}^{n-1}\left(\sum_{i=k+1}^{n}a_{i}\alpha^{i-k-1}\right)x^{k} $$
$$ =\left(a_{1}+a_{2}\alpha+a_{3}\alpha^{2}+a_{4}\alpha^{3}+\ldots+a_{n}\alpha^{n-1}\right) $$
$$ +\left(a_{2}+a_{3}\alpha+a_{4}\alpha^{2}+a_{5}\alpha^{3}+\ldots+a_{n}\alpha^{n-2}\right)x $$
$$ +\left(a_{3}+a_{4}\alpha+a_{5}\alpha^{2}+a_{6}\alpha^{3}+\ldots+a_{n}\alpha^{n-3}\right)x^{2} $$
$$ +\dots+a_{n}x^{n-1} $$
$$ =a_{1}+a_{2}(\alpha+x)+a_{3}\left(\alpha^{2}+\alpha x+x^{2}\right) $$
$$ +a_{4}\left(\alpha^{3}+\alpha^{2}x+\alpha x^{2}+x^{3}\right) $$
$$ +\ldots+a_{i}\left(\alpha^{i-1}+\alpha^{i-2}x+\ldots+\alpha x^{i-2}+x^{i-1}\right) $$
$$ +\ldots+a_{i}\left(\alpha^{n-1}+\alpha^{n-2}x+\ldots+\alpha x^{n-2}+x^{n-1}\right). $$
Multiplying by $\left(x-\alpha\right)$ and canceling like terms of opposite sign produces the desired result
$$ \left(x-\alpha\right)f_{1}\left[x\right]=a_{1}\left(x-\alpha\right)+a_{2}(\alpha^{2}-x^{2}) $$
$$ +a_{3}\left(\alpha^{2}x+\alpha x^{2}+x^{3}-\alpha^{3}-\alpha^{2}x-\alpha x^{2}\right) $$
$$ +\ldots+a_{i}\left(\alpha^{i-1}x+\alpha^{i-2}x^{2}+\ldots+\alpha^{2}x^{i-2}+\alpha x^{i-1}+x^{i}\right) $$
$$ -a_{i}\left(\alpha^{i}+\alpha^{i-1}x+\alpha^{i-2}x^{2}+\ldots+\alpha^{2}x^{i-2}+\alpha x^{i-1}\right) $$
$$ +\ldots+a_{n}\left(\alpha^{n-1}x+\alpha^{n-2}x^{2}+\ldots+\alpha^{2}x^{n-2}+\alpha x^{n-1}+x^{n}\right) $$
$$ -a_{n}\left(\alpha^{n}+\alpha^{n-1}x+\alpha^{n-2}x^{2}+\ldots+\alpha^{2}x^{n-2}+\alpha x^{n-1}\right) $$
$$ =f\left[x\right]. $$
Well, \begin{align*} f[x]-f[\alpha]&=\sum_{i=0}^na_ix^i-\sum_{i=0}^na_i\alpha^i \\ &=\sum_{i=0}^na_i(x^i-\alpha^i) \\ &=\sum_{i=0}^na_i(x-\alpha)\sum_{k=0}^{i-1}x^{k}\alpha^{i-k-1} \\ &=(x-\alpha)\sum_{i=0}^n\sum_{k=0}^{i-1}a_i\alpha^{i-k-1} x^{k}. \end{align*}
Now we interchange the two sums to get \begin{align*} f[x]-f[\alpha]&=(x-\alpha)\sum_{k=0}^{n-1}\sum_{i=k+1}^{n}a_i\alpha^{i-k-1}x^{k} \\ &=(x-\alpha)\sum_{k=0}^{n-1}\bar{a}_kx^k \\ &=(x-\alpha)f_1[x]. \end{align*}