I'm trying to prove that the remainder of a $n$-th grade Taylor formula is $$R_n=\frac{f^{(n+1)}(\mu) (x-x_0)^{n+1}}{ (n+1)!}$$ where $\mu$ is a value between $x$ and the centre $x_0$.
For $n=1$ it follows from Lagrange's theorem.
$n \rightarrow n+1$) Suppose that $$f(x)-\sum_{i=1}^n \frac{f^{(i)}(x_0) (x-x_0)^{i}}{ (i)!}=\frac{f^{(n+1)}(\mu) (x-x_0)^{n+1}}{ (n+1)!}$$
then it we have:
$$R_{n+1}=f(x)-\sum_{i=1}^{n+1} \frac{f^{(i)}(x_0) (x-x_0)^{i}}{ (i)!}=\frac{f^{(n+1)}(\mu) (x-x_0)^{n+1}}{ (n+1)!}- \frac{f^{(n+1)}(x_0) (x-x_0)^{n+1}}{ (n+1)!}$$
Again, by Lagrange's theorem we have $$R_{n+1}=\frac{f^{(n+2)}(\lambda)(\mu-x_0)(x-x_0)^{n+1}}{(n+1)!}$$ where $\lambda$ is between $\mu$ and $x_0$.
Is this part correct? How can I conclude the proof? Thank you!