Potentially Stably Constant

This proof might be incorrect, but I want to know what you think.

Let $f$ be a function satisfying the given conditions. Let $N := \min \{n \in \mathbb{N} \, | \, \forall x \in [0,1]: f^{(n)}(x) = 0\}$ . This minimum exists, because of the second condition and the fact that for all $x \in [0,1]$ and all $n,m \in \mathbb{N}: f^{(n)}(x) = 0 \implies f^{(n+m)}(x) = 0$ . Because of the choice of $N$ , it follows that $f^{(N)}$ is identically equal to 0 on the interval $[0,1]$ while $f^{(N-1)}$ is not identically 0 on the same interval. So, $f(x) = \underbrace{\int \cdots \int}_{N \text{ times}} f^{(N)}(x) \, \underbrace{\mathrm{d}x \cdots \mathrm{d}x}_{N \text{ times}} = \underbrace{\int \cdots \int}_{N \text{ times}} 0 \, \underbrace{\mathrm{d}x \cdots \mathrm{d}x}_{N \text{ times}} = C_{N-1} \cdot x^{N-1} + C_{N-2} \cdot x^{N-2} + \cdots + C_1 \cdot x + C_0,$ where $C_0, \ldots, C_{N-1}$ denote constant coefficients. We conclude that $f$ is a polynomial.