Prove a fomular $\sum_{j=0}^n(x_j-x)^kl_j(x)\equiv 0$ for $l_j(x)$ , the Lagrange Interplotion Basis $$ l_{j}(x)=\frac{\left(x-x_{0}\right) \cdots\left(x-x_{j-1}\right)\left(x-x_{j+1}\right) \cdots\left(x-x_{n}\right)}{\left(x_{j}-x_{0}\right) \cdots\left(x_{j}-x_{j-1}\right)\left(x_{j}-x_{j+1}\right) \cdots\left(x_{j}-x_{n}\right)} $$ $x_j$ are different points $(j=0,1,2,\cdots,n)$
My works: I have known the fomular $$ \sum_{j=0}^{n} x_{j}^{k} l_{j}(x) \equiv x^{k}, \quad k=0,1, \cdots, n $$ I want to prove $\sum_{j=0}^n(x_j-x)^kl_j(x)\equiv 0$
I use the above fomular to get the following for $k = 1,\cdots,n$ $$ \begin{aligned} & \sum_{j=0}^{n} \sum_{l=0}^{k}\left(\begin{array}{l} k \\ l \end{array}\right) x_{j}^{l}(-x)^{k-l} l_{j}(x) \\ =& \sum_{l=0}^{k}\left(\begin{array}{l} k \\ l \end{array}\right)(-x)^{k-l} \sum_{j=0}^{n} x_{j}^{l} l_{j}(x) \\ =& \sum_{l=0}^{k}\left(\begin{array}{l} k \\ l \end{array}\right)(-x)^{k-l} x^{l} \\ =&\,(x-x)^{k}\\ \equiv & \,0 \end{aligned} $$ I wonder is there a method doesn't use the above fomular, or can inspire me from a different opinion. Thanks!!!
My way:
If $f(x)$ is a $k$-order polynomial with $0\le k\le n$, it holds that $$ \sum_{j=0}^n f(x_j)l_j(x) \equiv f(x). $$ So, for every $t\in\mathbb{R}$, we can choose $f_t(x)=(x-t)^k$, then the equality $$ \sum_{j=0}^n (x_j-t)^k l_j(x) = (x-t)^k $$ holds for every $t$ and $x$.
If $k=1,2,\cdots,n$, let $t=x$, then $$ \sum_{j=0}^n (x_j-x)^k l_j(x) \equiv 0 $$ follows.