Sequence of antiderivatives of a continuous function

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function and $({f}_{n}{)}_{n\ge 0}$ a sequence of functions such that ${f}_{0}=f$ and ${f}_{n+1}$ is an antiderivative (primitive) of ${f}_{n}$ for each $n\ge 0$, with the property that for each $x\in \mathbb{R}$ there exists $n\in \mathbb{N}$ such that ${f}_{n}(x)=0$. Prove that f is identically 0.

Honestly, I do not have a viable starting point; but the problem looks very nice. Obviously, ${f}_{n}^{(n)}=f$ and one can use the general formula for the solutions of this differential equation of order n, but...

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function and $({f}_{n}{)}_{n\ge 0}$ a sequence of functions such that ${f}_{0}=f$ and ${f}_{n+1}$ is an antiderivative (primitive) of ${f}_{n}$ for each $n\ge 0$, with the property that for each $x\in \mathbb{R}$ there exists $n\in \mathbb{N}$ such that ${f}_{n}(x)=0$. Prove that f is identically 0.

Honestly, I do not have a viable starting point; but the problem looks very nice. Obviously, ${f}_{n}^{(n)}=f$ and one can use the general formula for the solutions of this differential equation of order n, but...