You are trying to understand why the remainder of Lagrange interpolation polynomial has the form $$\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)(x-x_1)\cdots(x-x_n)$$
Me too!And I find it.Indeed,Lagrange's work is based on Newtons'work.Newton has his polynomial interpolation method called Newton's interpolation,and the remainder of Newton's interpolation is in the form of $$R_n(x)=f[x,x_0,x_1,\cdots,x_n]\prod_{i=0}^n(x-x_i)$$
You'd better google Newton's method and learn it.
Ok,It seems that you are not satisfied with my answer,because you do not respond.So I give an update to my answer to provide you all the details.Let's first see the differential mean value theorem,it is stated as follows:
Let $f(x)$ be a real valued function defined on $\mathbf{R}$,and $f(x)$ is second differentiable on $\mathbf{R}$,and its second order derivative on $\mathbf{R}$ is continuous.Then there exists $\xi\in (x_0,x_1)$ such that
$$f'(\xi)=\frac{f(x_1)-f(x_0)}{x_1-x_0}$$
We prove the differential mean value theorem by constructing a function
$$g(x)=f(x)-[\frac{f(x_1)-f(x_0)}{x_1-x_0}(x-b)+f(x_1)]$$
$g(a)=g(b)=0$,so we can apply Rolle's theorem once to get the differential mean value theorem.Its picture is shown below:
Now we see a deeper example.
Let $f(x)$ be a real valued function defined on $\mathbf{R}$,and $f(x)$ is second differentiable on $\mathbf{R}$,and its second derivative on $\mathbf{R}$ is continuous.There is a polynomial of degree 2 passing through the points $(x_0,f(x_0)),(x_1,f(x_1)),(x_2,f(x_2))$.This unique polynomial is in the form of
$$a_2x^2+a_1x+a_0$$
Now the picture becomes
(I have to admit that my drawing skill is bad,my picture is not very accurate)
Once again,we want to estimate the interpolation error
$$f(x)-(a_2x^2+a_1x+a_0)$$
$a_2x^2+a_1x+a_0$ is not a good form.Lagrange has his form,but that's also not good for our purpose.Newton has his form,called Newton's interpolation,that's exactly what we need.According to Newton's interpolation,the interpolation polynomial is
$$Q(x)=f(x_0)+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)$$where $$f[x_0,x_1]=\frac{f(x_0)-f(x_1)}{x_0-x_1}$$,$$f[x_0,x_1,x_2]=\frac{f[x_0,x_1]-f[x_1,x_2]}{x_0-x_2}$$Now we investigate the function
$$g(x)=f(x)-(f(x_0)+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1))$$
Why we investigate this function?Because this function is in the form of
$$f(x)-Q(x)$$
further more,it is easy to verify that
$$g(x_0)=g(x_1)=g(x_2)=0$$(This is because these points are exactly the intersection points of the function $f(x)$ and the polynomial $Q(x)$)So apply Rolle's theorem twice ,we can get that $$f''(\xi)=2!f[x_0,x_1,x_2]$$
Similary,it is easy to verify that
$$f^{(n)}(\xi)=n!f[x_0,x_1,\cdots,x_n]$$
Now go back to Lagrange's interpolstion polynomial,we just need to prove that
$$f(x)=P(x)+f[x_0,\cdots,x_n,x](x-x_0)(x-x_1)\cdots (x-x_n)$$
This is just Newton's formula!Done.
To show that every $x_i$ is a root to $W$, we can re-write the definition of $W$ as
$$W(x) = Q(x) - P(x)$$ but $Q$ and $P$ are equal for all $x_i \ i = 1, 2, \ldots, n$ as they both interpolate the same $n$ points, so the claim follows.
For the degree of $W$, we use the fact that two polynomials are equal if and only if their coefficients corresponding to the same powers of $x$ are equal. As the left-hand side of $Q = P + W$ is a polynomial of degree $n$, the right hand-side must be, too. We know that $P$ is of degree $n-1$ so $W$ must have a $x^n$ term and no higher powers of $x$, proving the claim.
For the explicit formula for $W$, we can use the fundamental theorem of algebra and the fact that we know that $W$ is of degree $n$ and has $n$ real roots.
Best Answer
I have solved this problem myself.
In the general case: Suppose we are given the values of $n$ derivatives at a given point $x=a$ and $m$ derivatives at a point $x=b$. That is, $$P(a) = k_1,P^{'}(a)=k_2, \dots, P^{(i)}(a)=k_i,\dots, P^{(n-1)}(a) = k_n$$ $$P(b) = l_1,P^{'}(b)=l_2, \dots, P^{(i)}(b)=l_i,\dots, P^{(n-1)}(b) = l_m$$
We have the Interpolation formula: $$P(x) = \frac{(x-a)^n}{(b-a)^n}\left( l_1 + l_2(x-a) + l_3 \frac{(x-a)^2}{2!} + \dots \right) + \frac{(x-c)^m}{(a-b)^m}\left( k_1 + k_2(x-a) + k_3 \frac{(x-a)^2}{2!} + \dots \right)$$
Where the terms inside the parentheses are the first $n$ and $m$ terms of the Taylor series respectively.