Show, that $ \displaystyle \sum_{k=0}^n \lambda_k(0)x_k^j = \begin{cases} 1 \ (j=0)\\0 \ (j=1,2,\ldots n)\end{cases} $ Show, that $ \displaystyle \sum_{k=0}^n \lambda_k(0)x_k^j = \begin{cases} 1 \ (j=0)\\0 \ (j=1,2,\ldots n)\end{cases}$, where $\lambda_{k}$ is the 'helper' polynomial from Langrange Interpolation polynomial, with given $n+1$ points $x_{0},...,x_{n}$ etc.
So basically it's easily provable, that 
$\sum_{i=0}^{n}\left(x_{i}^{n}\lambda_{k}(x)\right)=x^n$. So what we've got here is $\sum_{k=0}^n \lambda_k(0)x_k^j=0^{j}$
So for $1\leq j$ it's clear that it's $1$, but what about $j=0$? $0^{0}$ is undefined, and I don't know what to do...
 A: Note that
If $f$ is $n + 1$ times continuously differentiable on a closed interval $I$ then there exists $\xi_x \in I$ such that
\begin{align}
f(x) - p_n(x) = \frac{f^{(n+1)}(\xi_x)}{(n+1)!}w_{n+1}(x)
\end{align}
For $j=0$:
\begin{align}
\sum_{k=0}^{n}\lambda_k(0) = \sum_{k=0}^{n}1 \cdot \lambda_k(0)
\end{align}
we interpolate the constant $1$-function, and hence
\begin{align}
1 - \sum_{k=0}^{n}\lambda_k(0) = \frac{0}{(n+1)!}w_{n+1}(x) = 0 \\
\Rightarrow \sum_{k=0}^{n}\lambda_k(0) = 1.
\end{align}
For $j=1, \dots, n$:
\begin{align}
\sum_{k=0}^{n}x_k^j \cdot \lambda_k(0)
\end{align}
hence for $x=0$:
\begin{align}
x^j - \sum_{k=0}^{n}\lambda_k(0)x_k^j =0 - \sum_{k=0}^{n}\lambda_k(0)x_k^j = \frac{f^{(n+1)}x^j}{(n+1)!}w_{n+1}(x) = 0 \\
\Rightarrow \sum_{k=0}^{n}\lambda_k(0)x_k^j = 0.
\end{align}
This of course doesn't work for $j=n+1$. In that case we have $f(x):=x^{n+1}$ and so $f^{(n+1)}(x)=(n+1)! \neq 0$. One can show that
\begin{align}
\sum_{k=0}^{n}\lambda_k(0)x_k^{n+1} = (-1)^nx_0 \dots x_n
\end{align}
for $x=0$.
