Bézout's identity on a polynome sequence I'm stuck on an exercise which is split in 3 questions :

1) Prove that : 
  $$\exists (U_n, V_n) \in \mathbb{R}[X]^2 \text{ s.t. } (1-X)^{n+1}U_n+X^{n+1}V_n=1$$
  2) Let $(R_n, S_n)\in\mathbb{R}[X]^2$ the remainders of the ED of respectively $U_n$ by $X^{n+1}$ and $V_n$ by $(1-X)^{n+1}$. Prove that :
  $$(1-X)^{n+1}R_n+X^{n+1}S_n=1$$
  3) Prove then that :
  $$\exists !(A_n,B_n)\in\mathbb{R}[X]^2, d°(A_n) \leq n \text{ and }d°(B_n)\leq n, \text{ s.t. } (1-X)^{n+1}A_n+X^{n+1}B_n=1$$

So for 1), I've been saying :
As $1$ is a root of multiplicity $n+1$ of $(1-X)^{n+1}$ and $d°((1-X)^{n+1})=n+1$, whereas $0$ is a root of multiplicity $n+1$ of $X^{n+1}$ and $d°(X^{n+1})=n+1$, $(1-X)^{n+1}$ and $X^{n+1}$ share no complex roots, which means :
$$X^{n+1} \wedge (1-X)^{n+1} = 1$$
Hence, using Bézout's identity :
$$\exists (U_n, V_n) \in \mathbb{R}[X]^2 \text{ s.t. } (1-X)^{n+1}U_n+X^{n+1}V_n=1$$
For the second question, I've been trying to write down the definition of $R_n$ and $S_n$ :
$$U_n = P_nX^{n+1}+R_n$$
$$V_n = Q_n(1-X)^{n+1}+S_n$$
And then trying multiples things but it leaded me nowhere.
Could you give me some hints ? (No complete proof please)
 A: 1 It is a proof devoid of content to invoke Bezout's to solve an exercise that is asking to prove Bezout's. I would instead use Euclid's algorithm to show how to construct some $U_n,V_n$.

Alternatively we can do this trick. Observe that $$\frac{U_n(X)}{X^{n+1}}+\frac{V_n(X)}{(1-X)^{n+1}}=\frac{1}{X^{n+1}(1-X)^{n+1}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(1)}$$
If we write $$U_n(X)=a_0+a_1X+...a_nX^n$$ and $$V_n(X)=b_0-b_1(X-1)+...+(-1)^nb_n(X-1)^n$$
We see that (1) is the partial fraction decomposition of $\frac{1}{X^n(1-X)^n}$. So we can find their coefficients directly. Since
$$U_n(X)+\frac{X^{n+1}V_n(X)}{(1-X)^{n+1}}=\frac{1}{(1-X)^{n+1}}$$ then $U_n(X)$ is the first $n+1$ terms of the Taylor expansion of $\frac{1}{(1-X)^{n+1}}$. But we know the Taylor expansion from the binomial theorem
$$\frac{1}{(1-X)^{n+1}}=(1-X)^{-(n+1)}\\=\sum_{k=0}^{\infty}(-1)^n\binom{-(n+1)}{k}X^k$$
So, $$\begin{align}a_k&=(-1)^k\binom{-(n+1)}{k}\\&=(-1)^k\frac{(-n-1)(-n-1-1)(-n-1-2)...(-n-1-(k-1))}{k!}\\&=\frac{(n+1)(n+2)...(n+k)}{k!}\end{align}$$
Similarly one can find $b_k$.

2 Assume you already have some equation $$(1-X)^{n+1}U_n+X^{n+1}V_n=1$$ and $U_n=X^nP+R$ with $d(R)\leq n$. Then $$(1-X)^{n+1}R+X^{n+1}[V_n+X^{n+1}P]=1$$
The claim is that we are done. You just need to count the degree of $(1-X)^{n+1}R$, notice that this must cancel with $X^{n+1}[V+X^{n+1}P]$, and deduce that $d[V+(1-X)^{n+1}P]\leq n$. Therefore, $V+(1-X)^{n+1}P$ is already the remainder of $V$ after division by $(1-X)^{n+1}$.
3 Follows from 2.
