Indian Statistical Institute B.Math & B.Stat : Differentiation

Let $f: \mathbb{R} \to \mathbb{R}$ be a function that is differentiable $n+1$ times for some positive integer $n$. The $ith$ derivative of $f$ is denoted by $f^{(i)}$. Suppose $$f(1)=f(0)=f^{(1)}(0)=\dots=f^{(n)}(0)=0.$$ Prove that $f^{(n+1)}=0$ for some $x \in (0,1)$. Since $f$ is $n+1$ times differentiable(everywhere in $\mathbb{R}$).$f,f^{(1)},\dots,f^{(n)}$ are all differentiable and continuous in $(0,1) \quad and \quad [0,1]$ respectively.$\dots (A)$ Since $f(0)=f(1)=0$ By Rolles' Theorem $\exists \quad c_1$ in $(0,1)$ suct that $f^{(1)}(c_1)=0$ Now consider the interval $[0,c_1]$. Note that $f^{(1)}(0)=f^{(1)}(c_1)=0$ and $f^{(1)}$ is continuous and differentiable in $[0,c_1]$ and $(0,c_1)$ respectively. (Using $\dots (A)$). Thus by Rolle's Theorem $\exists \quad c_2$ in $(0,c_1)$ suct that $f^{(2)}(c_2)=0$ Now consider the interval $[0,c_2]$. Note that $f^{(2)}(0)=f^{(2)}(c_2)=0$ and $f^{(2)}$ is continuous and differentiable in $[0,c_2]$ and $(0,c_2)$ respectively. (Using $\dots (A)$). Thus by Rolle's Theorem $\exists \quad c_3$ in $(0,c_2)$ suct that $f^{(3)}(c_3)=0$ Continuing like this, we get a point $c_{n}$ such that $f^{(n)}(c_{n})=0$ where $c_{n} \in (0,c_{n-1})$. Now consider the interval $[0,c_n]$. Note that $f^{(n)}(0)=f^{(n)}(c_n)=0$ and $f^{(n)}$ is continuous and differentiable in $[0,c_n]$ and $(0,c_n)$ respectively. (Using $\dots (A)$). Thus by Rolle's Theorem $\exists \quad c_{n+1}$ in $(0,c_n)$ suct that $f^{(n+1)}(c_{n+1})=0$. Since $0 < c_{n+1} < c_{n} < \dots < c_1 < 1$ thus $c_{n+1} \in (0,1)$