A bizarre proof of Taylor expansion Here is a problem from Gouvea's textbook on $p$-adic numbers.

Let $f(x) = a_0 + a_1x + \ldots + a_nx^n$ be a polynomial with coefficients in a field $K$ of characteristic zero. Show that the Taylor formula is true for $f(x)$, i.e., that for any $x, h \in K$, $$f(x+h) = f(x) + f'(x) h + \frac{1}{2!} f''(x)h^2 + \ldots$$

This can be done in many ways, usually by comparison of coefficients of $x^k$ on both sides of the equations, but author suggests a jazzier proof, which is an overkill. The field generated by $\mathbb Q$ and the coefficients of $f(x)$ can be embedded in $\mathbb C$, and the theorem is clearly true for polynomials with complex coefficients.
My questions are:


*

*How can I show existence of an embedding $\mathbb Q(a_0, a_1, \ldots, a_n) \to \mathbb C$?

*Why is the Taylor expansion true in $\mathbb C$?

 A: Every field $K$ of characteristic $0$ with transcendence degree $\operatorname{trdeg}(K|\mathbb Q) \leq \operatorname{trdeg}(\mathbb C|\mathbb Q) = 2^{\aleph_0}$ can be embedded into $\mathbb C$ as follows: Let $T$ be a transcendence basis for $K|\mathbb Q$ and $T'$ a transcendence basis for $\mathbb C|\mathbb Q$. Choose an injection $T \hookrightarrow T'$ to get an embedding $\mathbb Q(T) \hookrightarrow \mathbb Q(T')$. Since $K|\mathbb Q(T)$ is an algebraic extension and $\mathbb C$ is the algebraic closure of $\mathbb Q(T')$, the embedding can be extended to an embedding $K \hookrightarrow \mathbb C$.
As to why the Taylor expansion for polynomials is true in $\mathbb C$, I don't think there is an easier proof that would not work over an arbitrary field of characteristic $0$. It's just something that one would accept to be well-known from a course in analysis and not bother reproving it in a textbook an $p$-adic numbers. If you actually wanted to prove Taylor expansion for polynomials over $\mathbb C$, the easiest thing would still be to write 
$$f(x+h) = \sum_i p_i(x)h^i$$
with polynomials $p_i \in \mathbb C[x]$, take the $k$-fold partial derivative with respect to $h$ and compare the constant coefficient to get $p_k(x) = f^{(k)}(x)/k!$. But this works just as well over any characteristic $0$ field if you interpret "partial derivative" in the algebraic rather than analytic sense.
