$\sum_{j=i}^n\binom nj\binom jix^{n-j}=\binom ni\left(\frac1x+1\right)^{n-i}x^{n-i}$ Question :
$$\sum_{j=i}^n\binom nj\binom jix^{n-j}=\binom ni\left(\frac1x+1\right)^{n-i}x^{n-i}$$
I can't prove this identity. Any helps would be appreciated.
 A: $$
\begin{aligned}
\binom{n}{i}\left(\frac{1}{x}+1\right)^{n-i} x^{n-i} &=\sum_{l=0}^{n-i}\binom{n}{i}\binom{n-i}{l}x^{n-i-l}\\
&=\sum_{j=i}^n \binom{n}{i}\binom{n-i}{j-i}x^{n-j}\\
&=\sum_{j=i}^n \frac{n!}{i! (n-j)! (j-i)!} x^{n-j}\\
&=\sum_{j=i}^n \binom{n}{j}\binom{j}{i} x^{n-j}
\end{aligned}
$$
(I leave you the task of filling in the details :))
A: Re-write the right hand side as
$$\binom{n}{i}\left(\frac{1}{x}+1\right)^{n-i}x^{n-i}=\binom{n}{i}\left(1+x\right)^{n-i}$$
then using the "find the coefficient of $y^i$" operator $\left[y^i\right]$ the above is clearly
$$\left[y^i\right]\left(y+(1+x)\right)^n$$
which may alternatively be written
$$\begin{align}
\left[y^i\right]\left((1+y)+x\right)^n&=\left[y^i\right]\sum_{j=0}^{n}\binom{n}{j}x^{n-j}\left(1+y\right)^j\\[1ex]
&=\left[y^i\right]\sum_{j=0}^{n}\binom{n}{j}x^{n-j}\sum_{i=0}^{j}\binom{j}{i}y^i\\[1ex]
&=\left[y^i\right]\sum_{j=0}^{n}\sum_{i=0}^{j}\binom{n}{j}\binom{j}{i}x^{n-j}y^i\\[1ex]
&=\left[y^i\right]\sum_{i=0}^{n}\sum_{j=i}^{n}\binom{n}{j}\binom{j}{i}x^{n-j}y^i\\[1ex]
&=\left[y^i\right]\sum_{i=0}^{n}y^i\left(\sum_{j=i}^{n}\binom{n}{j}\binom{j}{i}x^{n-j}\right)\\[1ex]
&=\sum_{j=i}^{n}\binom{n}{j}\binom{j}{i}x^{n-j}\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare\end{align}$$
A: As your question currently stands, note that proving what you want is equivalent to proving $\displaystyle \sum_{j=i}^{n}\binom{n}{j}\binom{j}{i}x^{n-j}= \binom{n}{i}(1+x)^{n-i}$
Also note that both the LHS and RHS are polynomials in $x$ of degree $n-i$, so if we want to prove the above $\forall \, x \in \mathbb{R}$, it suffices to prove it $\forall \,  x \in \mathbb{N}$ since two polynomials each of degree $d$ agreeing on at least $d+1$ values, agree everywhere.
Let $x \in \mathbb{N}$ and let $[x]$ denote the set $\{1,2 \ldots, x\}.$ Here's a combinatorial proof via double-counting.
The RHS counts the number of ways of choosing a core committee of $i$ members out of $n$ people while labeling the $n-i$ non-core committee members with a label from $[x+1].$ This can be done in $\displaystyle \binom{n}{i}(x+1)^{n-i}$  ways.
Alternatively, for a fixed $j, \, i\leq j \leq n,$ from a group of $n$ people, first choose a $j$-member committee in $\displaystyle \binom{n}{j}$ ways and from this committee choose an $i$-member core committee in $\displaystyle \binom{j}{i}$ ways such that the $n-j$ non-committee members get a label from $[x]$ in $x^{n-j}$ ways and the $j-i$ committee members who are $\textbf{not}$ on the core committee get the label $(x+1).$ This can be done in $\displaystyle \binom{n}{j}\binom{j}{i}x^{n-j}$ ways.
Summing over $j = i, i +1, \ldots, n$, covers all the possible strengths of the intermediate committee from which an $i-$member core committee can be chosen and at each value of $j,$ exactly $n-j+j-i = n-i$ people have a label from $[x+1]$, like in the RHS scenario. 
The desired identity follows.
