Putnam 2018 - Exercise A.5 - proof check

The problem statement is as follows.

Let $f:Bbb R to Bbb R$ be an infinitely differentiable function satisfying $f(0) = 0$ and $f(1) = 1$, and $f(x) geq 0$ for all $x in Bbb R$. Show that there exist a positive integer $n$ and a real number $x$ such that $f^{(n)}(x)< 0$.

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My proof.

Suppose $f^{(n)}(x) geq 0$ for all $xin Bbb R$, for all $n in Bbb Z^+$.

If there exists a $c < 0$ such that $f(c)>0$, then the Mean Value Theorem would give a negative derivative at some point between $c$ and $0$. So $f(x) = 0$ for $x leq 0$, and thus, since for all $n$ $$lim_{xto 0^-}f^{(n)}(x) = 0.$$
we must have $f^{(n)}(0) = 0$. (So far it is all like in some of the "official" proofs.)

By Taylor’s Theorem (with "starting point" $0$) we have
$$f(x) = frac{f^{(n)}(eta_n(x))}{n!}x^n,$$
for some $eta_n(x) in (0,x)$. In particular
$$f(1) = frac{f^{(n)}(eta_n(1))}{n!} = 1,$$
which implies that there is a point $eta_n(1)in (0,1)$ such that $f^{(n)}(eta_n(1)) = n!$. By monotonicity of $f^{(n)}(x)$, therefore we must have
$$f^{(n)}(1) geq n!.$$
Using again Taylor’s Theorem (starting at $1$, this time) yields
$$f(x) = sum_{k=0}^n frac{f^{(k)}(1)}{k!}(x-1)^k+ frac{f^{(n+1)}(xi_n(x))}{(n+1)!}(x-1)^{n+1},$$
for some $xi_n(x) in (1,x)$, and where $f^{(0)}(x) = f(x)$.
This gives
$$f(2) geq n$$
for all $nin Bbb Z^+$, a contradiction.

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Is my proof correct?