In this proof, we showed the Lagrange's form of the remainder:
\[ \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1}= R_n(x)\]
Let there be some positive integer \(M\) such that \(|f^{(n+1)}(x)| \le M\) for all \(x \in I\) (including \(c\)):
\[\begin{gather} |f^{(n+1)}(c)| \le M \\ \frac{|f^{(n+1)}(c)|}{(n+1)!} (x-a)^{n+1} \le \frac{M}{(n+1)!} (x-a)^{n+1} \end{gather}\]
By definition of \(R_n(x)\):
\[ |R_n(x)| \le \frac{M}{(n+1)!} (x-a)^{n+1}\]