Derivation of Taylor expansion with the $a$ term If we have
$$f(x) = \sum_{n=0}^\infty a_n x^n$$
The $k$th derivative is 
$$f^{(k)}(x) = \sum_{n=0}^{\infty} a_{n+k} \frac{(n+k)!}{n!} x^n$$
Which also means that
$$f^{(k)}(0) = k! a_k$$
Implying
$$a_k = \frac{f^{(k)}(0)}{k!}$$
So we can substitute this back into the original definition for $f(x)$:
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$
And this shows the Taylor series expansion.
However I've also noticed this other definition:
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n$$
But I can't seem to use the previous logic to get the math to work out when I use $x=a$ instead of $x=0$, since:
$$f^{(k)}(a) = \sum_{n=0}^{\infty} a_{n+k} \frac{(n+k)!}{n!} a^n$$
But from here it doesn't exactly condense nicely and so I don't know how to substitute it to make it work out properly.

Same logic still applies
$f(x) = \sum_{n=0}^{\infty} a_{n} (x-a)^n\\
f^{(k)}(x) = \sum_{n=0}^{\infty} a_{n+k} \frac{(n+k)!}{n!} (x-a)^n\\
f^{(k)}(a) = a_{k} k!\\
a_k = \frac {f^{(k)}(a)}{k!}$
