Let $ F $ be a field and consider the polynomial ring $ F[x] $. I know that $ F[x] $ is a principal ideal domain and so the ideal $\langle x^{n - 2} - x^{n - 1}, x^{n - 3} - x^{n - 1}, \dots, x - x^{n - 1}, 1 - x^{n - 1} \rangle $ is generated by a single element in $ F[x] $. My guess is that $$ \langle x^{n - 2} - x^{n - 1}, x^{n - 3} - x^{n - 1}, \dots, x - x^{n - 1}, 1 - x^{n - 1} \rangle = \langle 1 - x \rangle. $$
I have proven that $ \langle x^{n - 2} - x^{n - 1}, x^{n - 3} - x^{n - 1}, \dots, x - x^{n - 1}, 1 - x^{n - 1} \rangle \subset \langle 1 - x \rangle $, but I am struggling to prove the other direction.
My approach is to write an element $ f(x) = \sum_{i = 0} a_{i} x^{i} $ in $ \langle 1 - x \rangle $ in the form $ f_{1}(x^{n - 2} - x^{n - 1}) + f_{2}(x^{n - 3} - x^{n - 1}) + \dots + f_{n - 2}(x - x^{n - 1}) + f_{n - 1}(1 - x^{n - 1}) $ for $ f_{i} \in F[x] $.
If the degree of $ f $ does not exceed $ n - 2 $, then I can construct $$ f(x) = a_{0}(1 - x^{n - 1}) + (a_{1} - a_{0})(x - x^{n - 1}) + (a_{2} - a_{1})(x^2 - x^{n - 1}) + \dots + (a_{n - 3} - a_{n - 4})(x^{n - 3} - x^{n - 1}) + (a_{n - 2} - a_{n - 3})(x^{n - 2} - x^{n - 1}).$$ This implies $ f(x) \in \langle x^{n - 2} - x^{n - 1}, x^{n - 3} - x^{n - 1}, \dots, x - x^{n - 1}, 1 - x^{n - 1} \rangle $, but this fails when $ \deg(f) > n - 2 $.
We know that every element in the ideal $\langle 1-x\rangle$ is a multiple of $(1-x)$, so:
$$f(x) = (1-x) \sum_{i = 0}^m a_{i} x^{i}$$
Generate $1-x$ simply by subtracting the second-to-last element in the basis from the last:
$$(1-x^{n-1}) - (x-x^{n-1}) = 1-x$$
We're mostly done:
$$f(x) = [(1-x^{n-1}) - (x-x^{n-1})] \sum_{i = 0}^m a_{i} x^{i}$$
Just set the other coefficients to zero.