Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$? Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$?
Can someone point me to a proof, or explain if it's a simple answer?
What I'm looking for is the point where it becomes understood that trigonometric functions and pi can be expressed as series.  A lot of the information I find when looking for that seems to point back to arctan.

The derivative of the arc tangent is
$$\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}.$$
From the formula for geometric series (see for example this answer for a proof) shows that
$$1+y+y^2+y^3+\cdots = \frac{1}{1-y}\qquad\text{if }|y|\lt 1.$$
Plugging in $-x^2$ for $y$, we get that 
$$\begin{align*}
\frac{1}{1+x^2} &= \frac{1}{1-(-x^2)} \\
&= 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \cdots + (-x^2)^n + \cdots\\
&= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots
\end{align*}$$
provided that $|-x^2| \lt 1$; that is, provided $|x|\lt 1$. All the computations below are done under this hypothesis (see comments at the end).
So we have that:
$$\frac{d}{dx}\arctan(x) = 1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\qquad\text{if }|x|\lt 1$$
Because this is a Taylor series, it can be integrated term by term. That is, up to a constant, we have:
$$\begin{align*}
\arctan(x) &= \int\left(\frac{d}{dx}\arctan (x)\right)\,dx \\
&= \int\left(1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\right)\,dx\\
&= \int\left(\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\right)\,dx\\
&= \sum_{n=0}^{\infty}\left(\int (-1)^{n}x^{2n}\,dx\right)\\
&= \sum_{n=0}^{\infty}\left((-1)^{n}\int x^{2n}\,dx\right)\\
&= \sum_{n=0}^{\infty}\left((-1)^{n}\frac{x^{2n+1}}{2n+1}\right) + C\\
&= C + \left( x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} +\cdots\right).
\end{align*}$$
Evaluating at $x=0$ gives $0 = \arctan(0) = C$, so we get
$$\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} + \cdots,\qquad\text{if }|x|\lt 1.$$
the equality you ask about.
Note however that this does not hold for all $x$: it certainly works if $|x|\lt 1$, by the general properties of Taylor series. But the arc tangent is defined for all real numbers. The series we have here is
$$\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1}.$$
Using the 
Ratio Test, we have that
$$\begin{align*}
\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} &= \lim_{n\to\infty}\frac{\quad\frac{|x|^{2n+3}}{2n+3}\quad}{\frac{|x|^{2n+1}}{2n+1}}\\
&= \lim_{n\to\infty}\frac{(2n+1)|x|^{2n+3}}{(2n+3)|x|^{2n+1}}\\
&= \lim_{n\to\infty}\frac{|x|^2(2n+1)}{2n+3}\\
&= |x|^2\lim_{n\to\infty}\frac{2n+1}{2n+3}\\
&= |x|^2.
\end{align*}$$
By the Ratio Test, the series converges absolutely if $|x|^2\lt 1$ (that is, if $|x|\lt 1$) and diverges if $|x|\gt 1$. At $x=1$ and $x=-1$, the series is known to converge. So the radius of convergence is $1$, and the equality is valid for $x\in [-1,1]$ only (that is, if $|x|\leq 1$; we gained two points in the process). 
However, the arc tangent has a nice property, namely that 
$$\arctan\left(\frac{1}{x}\right) = \frac{\pi}{2} - \arctan(x),$$
So, given a value of $x$ with $|x|\gt 1$, you can use this identity to compute $\arctan(x)$ by computing $\arctan(\frac{1}{x})$ instead, and for this argument the series is valid.
