Let's start with a simpler task:
We wish to approximate the function $f$ with simpler functions, namely polynomial functions. In fact, we will construct an approximation scheme that gives a sequence of polynomial functions that approximate $f$. The sequence will be denoted by $P_0(x)$, $P_1(x)$, $P_2(x)$, $\ldots\,$, where each $P_n(x)$ is a polynomial of degree $n$. Moreover, the scheme will be simple in the sense that, given $P_n$ and $P_m$ with $n, the coefficients $x^i$ are identical for $i\le n$.
We decide to define $P_n(x)$ to be the polynomial of degree $n$ that has the same value as $f$ at $x=0$ and has the same derivatives as $f$ at $x=0$ up to the $n^{\rm th}$-derivative. That is, we demand
$P_n(0)=f(0),\quad\text { and }\quad P_n^{(k)}(0) = f^{(k)}(0) \text{ for }k=1, 2, \ldots\,, n.$
So, $P_n(x)$ has the form $ P_n(x) = a_0+ a_1x+ a_2x^2+ a_3x^3+a_4x^4+\cdots+a_n x^n; $ where $ P_n^{(k)}(0) = f^{(k)}(0) \quad{\rm for}\quad k=1, 2, \ldots\,, n. $ We can solve for the unknown constants without too much difficulty. We first write down the derivatives of $P_n(x)$:
$ \eqalign{ P_n(x) &= a_0+ a_1x+ a_2x^2+ a_3x^3+a_4x^4+\cdots+a_n x^n \cr P_n'(x) &= a_1+ 2a_2x+ 3a_3x^2+4 a_4x^3+\cdots+n a_n x^{n-1} \cr P_n''(x) &= 2a_2+ 3\cdot2 a_3x+4\cdot3 a_4x^2+\cdots+n(n-1) a_n x^{n-2} \cr P_n'''(x) &= 3\cdot2a_3+ 4\cdot3\cdot2a_4x+\cdots+n(n-1)(n-2) a_n x^{n-3} \cr &\ \ \vdots\cr P_n^{(n)}(x) &= n! a_n \cr } $ Now, we set $f^{(k)}(0) = P_n^{(k)}(0)$ and solve for the unknown coefficients: $ \eqalign{ f(0)=P_n(0)&=a_0\cr f'(0)= P'_n(0)&=a_1\cr f''(0)=P_n''(0)&=2! a_2\cr f'''(0) = P_n'''(0)&=3!a_3\cr &\vdots \cr f^{(n)}(0)=P_n^{(n)}(0)&=n! a_n\cr } \eqalign{ &\Rightarrow\vphantom{f v(0)}\cr &\Rightarrow\vphantom{f'(0)}\cr &\Rightarrow\vphantom{f'(0)}\cr &\Rightarrow\vphantom{f'(0)}\cr \ \ \ \ &\ \ \vdots\cr &\Rightarrow\vphantom{f'(0)}\cr } \eqalign{ a_0 &= f'(0) \cr a_1 &=f'(0)\cr a_2 &= f''(0)/2! \cr a_3 &=f'''(0)/3!\cr &\vdots \cr a_n &=f^{(n)}(0)/n! \cr } $ Thus, the Taylor Polynomial of degree $n$ for the function $f$ is
$\tag{1} P_n(x) = f(0)+{f'(0)\over 1!}x+{f''(0)\over 2!}x^2+{f'''(0)\over3!} x^3+\cdots+{f^{(n)}(0)\over n!}x^n $ Note that to compute a Taylor Polynomial, one need only find the values $\def\hfil{}{\hfil f(0),\ f'(0),\ f''(0),\ \ldots\,, \ f^{(n)}(0) \hfil} $
and substitute into formula (1).
Let's find the Taylor Polynomial for
$f(x)=e^x$:
Example: Find the Taylor Polynomial of degree $n$ for $f(x)=e^x$.
We first evaluate the derivatives of $e^x$ at $x=0$. Since ${d^k\over dx^k }e^x = e^x$ for any $k$, we have $f^{(k)}(0)=e^0=1$ for all $k$. Thus for any $n$: $ P_n(x) =1+x+{x^2\over 2!}+{x^3\over 3!}+{x^4\over 4!}+\cdots +{x^n\over n!}. $ In particular: $ \eqalign{ P_0(x)&= 1 \cr P_1(x)&= 1+x \cr P_2(x)&= 1 +x+{x^2\over2}\cr P_3(x)&= 1 +x+{x^2\over2}+{x^3\over6}.\cr } $ The graphs of $y=e^x$ and the Taylor Polynomials $P_0$, $P_1$, $P_2$, and $P_3$ are shown below: 
Note that the approximations are very good, for $x$ close to 0, once $n\ge2$. Also, the graphs of the Taylor Polynomials seem to be converging to the graph of $f$. However, once $x$ becomes big, the approximation for a fixed $P_n$ becomes bad.
What if we kept going? That is, what if we formed the infinite degree polynomial?
The power series you have is obtained by taking the Taylor polynomial for
$f(x)=e^x$ of "infinite degree".
That this can actually be done, and that the resulting series represents $f(x)=e^x$ takes a good deal of machinery. One needs the following results and definitions (whose proofs can be found in any Calculus text worthy of the name):
Fact 1: Suppose that $f$ is $n$-times continuously differentiable on the interval $[0,x]$ (or $[x,0]$) and that $f^{(n+1)}$ exists on $(0,x)$ (or $(x,0)$). Then $ f(x)=P_n(x) + R_n(x),$ where $P_n(x)$ is as in (1) and $ \tag{2}R_n(x)= {f^{(n+1)}(c)\over (n+1)! } x^{n+1} $ for some number $c$ between 0 and $x$.
Definition: The Taylor Series of $f$ is: $ P(x)= f(0)+{f'(0)\over 1!} x +{f''(0)\over 2!} x^2+{f'''(0)\over 3!} x^3+\cdots. $ Note that the Taylor Series of the function $f$ is an infinite series whose first $(n+1)$-terms give the Taylor Polynomial of degree $n$ of $f$.
Fact 2: The Taylor Series $P(x)$ of the function $f$ converges to the function value $f(x)$ if and only if the remainder term $R_n(x)$ for the Taylor Polynomial $P_n(x)$ of $f$ converges to zero. That is $ f(x) = \sum_{n=0}^\infty {f^{(n)}(0)\over n!} x^n, $ If and only if $ \lim_{n\rightarrow\infty} R_n(x) =0, $ where $R_n(x)$ is defined as in (2).
Now, the Taylor series for
$f(x)=e^x$ can be found as in the previous
Example. In fact, we have:
$ P(x)=\sum_{n=0}^\infty {x^n\over n!} $ Using Fact 2, one can show that the Taylor Series for $f$ indeed converges to $e^x$ for any $x$:
$ e^x=\sum_{n=0}^\infty {x^n\over n!}. $
(You just show that for fixed $x$, $R_n(x)$ has limit 0. Here, note that with $x$ fixed, one can bound the terms $|f^{(n+1)(c)|}$ above.)