Let $f:\mathbb{R} \to \mathbb{R}$ be an infinitely differentiable function and suppose that for some $n >1$, $$ f(1) = f(0) = f^{(1)}(0) = f^{(2)}(0) = \cdots = f^{(n)} (0) = 0 $$ where $f^{(k)}(x)$ denotes the $k$-th derivative of $f$ for $k \ge 1$.
Prove that there exists $x$ with $(0<x<1)$ such that $f^{(n+1)}(x) = 0$.
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By MVT, there is $c_1\in(0,1)$ such that $\displaystyle f^{(1)}(c_1)=\frac{f(1)-f(0)}{1-0}=0$.
By MVT again, there is $c_2\in(0,c_1)$ such that $\displaystyle f^{(2)}(c_2)=\frac{f^{(1)}(c_1)-f^{(1)}(0)}{c_1-0}=0$.
By MVT again, there is $c_3\in(0,c_2)$ such that $\displaystyle f^{(3)}(c_3)=\frac{f^{(2)}(c_2)-f^{(2)}(0)}{c_2-0}=0$.
Repeat the process, we can construct $c_1,c_2,\dots, c_{n+1}$ such that $f^{(k)}(c_k)=0$ for $k=1,2,\dots,n+1$ and $1>c_1>c_2>\cdots>c_{n+1}>0$.
Taylor's theorem $$ f(x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!} (x - x_0)^k + R_n(x) $$ and the Lagrange form of the remainder $$ R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x - x_0)^{n+1} $$ for some $\xi \in (x_0, x)$, applied at $x_0 = 0$ give $$ 0 = f(1) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} 1^k + R_n(1) = R_n(1) = \frac{f^{(n+1)}(\xi)}{(n+1)!} $$ for some $\xi \in (0, 1)$.