Proof of Lagrange Polynomial I am trying to prove the following concepts of the Lagrange Polynomial:


*

*$\sum_{j=0}^n L_j(x)=1$

*$\sum_{j=0}^n x_j^m(x)L_j(x)=x^m, m \le n $
This is my work so far, but I am a little stuck on finishing part 1 and I don't know how to complete part 2.
So, the Lagrange polynomial is defined by:
$P_n(x)=\sum_{j=0}^n L_j(x)f_j$
$L_j(x)={\prod{j=0, j\neq i}^n (x-x_j)\over \prod{j=0, j\neq i}^n (x_i-x_j)}$
And the Lagrange Polynomial is equal to:
$P_n(x) = L_0(x)y_0+...+L_n(x)y_n$
If you plug in the a corresponding $x$ you get that specific $y$ term:
$P_n(x_0) = 1*y_0+ 0*y_1+...+0*y_n = y_0$
$P_n(x_1) = 0*y_0+ 1*y_1+...+0*y_n = y_1$ 
I need to prove this for any $x_j$, for all cases.
I can use this equation, where $f_j$ is equal to the $y$ term:
$P_n(x)=\sum_{j=0}^n L_j(x)f_j$
And choose a specific function that will helpful in completing my proof, but I am stuck. 
If someone could help me finish the first and provide a comprehensive explanation for the second proof, it would be much appreciated, thank you.
 A: For the first one, consider $\Psi(x) = \sum_{j=0}^n L_j(x) -1$. What can you say about $\Psi$? (Hint: how many distinct roots does it have?)
For the second one, do the same thing.
A: For 1., isn't that equivalent to choosing $f_j=1$ for all j and then using the fact that L interpolates correctly? (i.e. because choosing $f_j=1$ for all j is just interpolating the constant function one.)
The same also seems to be true of the choice of coefficients for part 2. (i.e. you are interpolating the values of the function $x^m$ at the chosen points.)
So the proof just reduces to showing that Lagrange polynomials interpolate correctly (and in particular interpolate all polynomials of degree less than or equal to their own degree exactly/perfectly); 1 and 2 are just special cases of that.
Also as Ahmed points out below you probably have to show that the resulting polynomials also have the correct degree (i.e. number of zeros) if that doesn't follow already from Lagrange polynomials interpolating polynomials exactly.
That is why there is the condition $m \le n$ in 2; if you try to interpolate a cubic polynomial using two points, your Lagrange polynomial will be quadratic and hence unable to fit the curve exactly.
