# Existence of an infinitely differentiable function $f$ with ${f^{(n)}}(0) = 0$ for all $n \in \mathbb{N}$.

How can one show that there exists an infinitely differentiable function $f: \mathbb{R} \to \mathbb{R}$ such that ${f^{(n)}}(0) = 0$ but $f^{(n)} \not\equiv 0$ for all $n \in \mathbb{N}$?

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The function $f(x)=e^{-1/x^2}$ is a canonical such example.
The function $f$ given by $$f(x)= \begin{cases} \exp\left(-1/x\right),\quad&x>0,\\ 0,&x\leq 0 \end{cases}$$ has derivatives of all orders that satisfy $f^{(n)}(x)=0$ for $x\leq 0$ and $n\in\mathbb N$. See e.g. wikipedia and/or this question.