Application of Taylor's Theorem in Number Theory I'm working through Alan Baker's book A Concise Introduction to the Theory of Numbers, and there's an assertion in there that confuses me. Here's the quote:

It is easily seen that no polynomial $f(n)$ with integer coefficients can be prime for all $n$ in $\mathbb{N}$, or even for all sufficiently large $n$, unless $f$ is constant. Indeed, by Taylor's theorem, $f(mf(n)+n)$ is divisible by $f(n)$ for all $m$ in $\mathbb{N}$

How is this an application of Taylor's theorem? It's entirely mysterious to me. Thanks in advance for any insight on this.
 A: ${\rm mod}\ \color{#c00}{f(n)}\!:\,\ f(m\color{#c00}{f(n)}+n)\equiv f(\color{#c00}0+n)\equiv 0\,\ $ by the Polynomial Congruence Rule $\ \ $ QED

Remark $\ $ To explain the relationship to Taylor's Theorem, the above amounts to using only the first couple of terms of the Taylor series, and this amounts to using the Factor Theorem, namely
$$\begin{align} f(x)\, &=\, f(n)\, +\, (x\!-\!n)(f'(n) +\, \cdots),\ \ \ \text{Taylor series at }\ x=n\\[4pt]
\Rightarrow\quad\ \, f(x)\, &=\, f(n)\, +\, (x\!-\!n)\, g(x)\ \text{ for some }\ g(x)\in \Bbb Z[x]\\[4pt]
\Rightarrow\ \ m f(n)\, &=\, \underbrace{x\!-\!n\mid f(x)-f(n)}_{\rm Factor\ Theorem}\,\Rightarrow\, f(n)\mid f(x)\ \ {\rm for}\ \ x = mf(n)\!+\!n\end{align}\qquad$$
But, of course, it is a bit overkill to use Taylor's Theorem to derive the Factor Theorem. And the Factor Theorem is a special case of the Polynomial Congruence Rule just as above, i.e.
${\rm mod}\ x\!-\!n\!:\,\ \color{#c00}{x\equiv n}\ \Rightarrow\, f(\color{#c00}x)\equiv f(\color{#c00}n)\,\ $ by the Polynomial Congruence Rule.
See here for further discussion of the Factor Theorem from a congruence standpoint.
A: By Tayor's Theorem
$$f(x) = f(n) + f'(n)(x-n) + \frac{f''(n)}{2!}(x-n)^2 + \cdots + \frac{f^{(k)}(n)}{k!}(x-n)^k + h_k(x)(x-n)^k$$
Now take $k$ big enough so that $h_k(x)=0$ (possible since $f$ is a polynomial).
So 
$$f(x) = f(n) + f'(n)(x-n) + \frac{f''(n)}{2!}(x-n)^2 + \cdots + \frac{f^{(k)}(n)}{k!}(x-n)^k$$
Now plug in $mf(n)+n$ for $x$.  You get an $f(n)$ in every term.
