Where do the factorials come from in the Taylor series? Unfortunately, I don't have much detail to give here. But is the general idea to cancel out the constant obtained from taking the derivative.
For instance, say my function was $f(x)=f_0+f_1x+f_2x^2+\dotsb$
Then $f'(x)=f_1+2f_2x+\dotsb$.
And if the expansion is centered around $x=0$, then
\begin{align}f'(0)&=0 \\
f''(0)&=2f_2\\
f'''(0)&=3\cdot 2f_3.\\
\end{align}
Therefore
\begin{align}
f_0&=f(0) \\
f_1&=\frac{f'(0)}{1} \\
f_2&=\frac{f''(0)}{2}
\end{align}
And so forth. Is that where the factorial comes from?
It is quite clear for a polynomial, but what about a trig function such as $\sin(x)$ other than using Taylor's formula?
 A: Watch this video to see why there are factorials:
http://www.youtube.com/watch?v=QMJvRNFhEGc   This guy is a simple, but effective teacher.
In short, the Taylor Series expansion is derived from the Power Series formula.
Power Series formula is: $f(x) = a + ax^1 + ax^2 + ax^3 + ...$
For example, multiple derivatives of $f(0) = ax^5$ leads to:
$f(0) = ax^5$
$f '(0) = 5ax^4$ then...
$f ''(0) = 20ax^3$  but wait!!  Instead, write the second derivative as $5*4a(x^3)$
$f '''(0) = 5*4*3a(x^2) $
$f ''''(0) = 5*4*3*2a(x)$
$f '''''(0) = 5*4*3*2*1a$
Solve for $a$ yields $a = f '''''(0)/5!$  Now, insert this $a$ value for the Power Series term $ax^5$.  So, $[f '''''(0)/5!]x^5$
Not easy to edit with this but hopefully the video will help.
A: It sounds like you already accept that the $n!$ terms make sense when you're talking about polynomials. For other functions like $\sin{x}$, the whole motivation for Taylor series is to approximate those functions by polynomials, so in my opinion I would say that the $n!$ terms appear because that is precisely the property that mathematicians wanted out of Taylor series when they first invented it - so that any random function, $\sin{x}$, $\ln{x}$, etc, could look like a polynomial.
Alternatively, maybe this can help you see: if we have the Taylor series for $f(x)$ at $0$,
$$ f(0) + f'(0)x + \frac{1}{2} f''(0) x^2 + \frac{1}{3!} f'''(0) x^3 + \ldots$$
then if we differentiate this function once, we get
$$ f'(0) + f''(0) x + \frac{1}{2} f''(0) x^2 + \ldots $$
which gives us the Taylor series for $f'(x)$ at $0$! Notice that all the terms "shifted" downwards; allowing us to recover the familiar form of the Taylor series.
A: Start with the fundamental theorem of calculus:
$$
     f(x) = f(x_0) + \int_{x_0}^x f^\prime(y) \mathrm{d} y
$$
and reapply it to $f'(y)$:
$$
   f(x) = f(x_0) + \int_{x_0}^x \left( f^\prime(x_0) + \int_{x_0}^y f^{\prime\prime}(z) \mathrm{d} z  \right) \mathrm{d} y = f(x_0) +f^\prime(x_0) \int_{x_0}^x \mathrm{d} y + \underbrace{\int_{x_0}^x \left( \int_{x_0}^y f^{\prime\prime}(z) \mathrm{d} z\right)\mathrm{d} y}_{\mathcal{R}_2(x)}
$$
Repeat this with $f^{\prime\prime}(z)$:
$$
    f(x) = f(x_0) + f^\prime(x_0) \underbrace{\int_{x_0}^x \mathrm{d} y}_{I_1(x)}  + f^{\prime\prime}(x_0) \underbrace{\int_{x_0}^x \int_{x_0}^y \mathrm{d}z \mathrm{d} y}_{I_2(x)} + \underbrace{\int_{x_0}^x \int_{x_0}^y \int_{x_0}^z f(w) \mathrm{d} w \mathrm{d} z \mathrm{d} y}_{\mathcal{R}_3(x)}
$$
and by continuing, we get:
$$
  f(x) = f(x_0) + f^\prime(x_0) \int_{x_0}^x \mathrm{d} y + \cdots + f^{(k)}(x_0) \underbrace{\int_{x_0}^{x} \int_{x_0}^{y_1} \int_{x_0}^{y_2} \cdots \int_{x_0}^{y_{k-2}} \mathrm{d} y_{k-1} \cdots\mathrm{d} y_3 \mathrm{d} y_2 \mathrm{d} y_1}_{I_k(x)} + \mathcal{R}_{k+1}(x)
$$
The iterated integrals $I_k(x)$ are easy to evaluate. They can be defined recursively
$$
   I_0(x) = 1, \quad I_k(x) = \int_{x_0}^x I_{k-1}(y) \mathrm{d} y
$$
Giving $I_k(x) = \frac{1}{k!} (x-x_0)^k$.
A: Since other answers have already covered the "real" reason, let's try to make an "intuitive" explanation.
If you're trying to approximate the function $f(x)$ with a 3rd degree polinomial $P(x)=ax^3+bx^2+cx+d$ around the point $x=0$. The least you can ask is that their first, second and third derivative match at $x=0$. So:
$$f(0) = P(0) = d$$
$$f'(0) = P'(0) = c$$
$$f''(0) = P''(0) = 2b$$
$$f'''(0) = P'''(0) = 6a$$
Put otherwise:
$$d = f(0)$$
$$c = f'(0)$$
$$b = \frac{1}{2} f''(0)$$
$$a = \frac{1}{6} f'''(0)$$
Of course, the reasoning would be similar for a general $n$th degree polynomial $Q(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0$, where
$$f^{(k)}(0) = Q^{(k)}(0) = k! a_k $$
Therefore:
$$a_k = \frac{1}{k!} f^{(k)}(0)$$
Of course, what we've stated here still holds for Taylor polynomials not centered around $x=0$
A: $\forall x \in \mathbb{R}\int_0^xt dt = \frac{x^2}{2!}$
$\forall x \in \mathbb{R}\int_0^x\frac{t^2}{2!} dt = \frac{x^3}{3!}$
$\forall x \in \mathbb{R}\int_0^x\frac{t^3}{3!} dt = \frac{x^4}{4!}$
$\forall x \in \mathbb{R}\int_0^x\frac{t^4}{4!} dt = \frac{x^5}{5!}$
