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.

Marty Cohen's answer gives an explanation for the following fact:
$$
\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}
$$
Here is an alternative explanation.
Let $y = \arctan(x)$, and try to find $\frac{dy}{dx}$.  We have 
$$
y = \arctan(x)
$$
$$
\tan(y) = \tan(\arctan(x))
$$
We'd like to simplify the right-hand side.  The definition of $\arctan(x)$ is:
$$
\arctan(x) = \textrm{the angle between }-\frac{\pi}{2} \textrm{ and } \frac{\pi}{2}\textrm{ whose tangent is } x.
$$
So whatever $\arctan(x)$ is, its tangent must be $x$.  That is,
$$
\tan(\arctan(x)) = x.
$$
So we have
$$
\tan(y) = x
$$
Now differentiate both sides with respect to $x$ (this is called implicit differentiation):
$$
\frac{d}{dx} \tan(y) = \frac{d}{dx} x
$$
$$
\sec^2(y) \frac{dy}{dx} = 1
$$
$$
\frac{dy}{dx} = \cos^2(y)
$$
$$
\frac{d}{dx}\arctan(x) = \cos^2(\arctan(x))
$$
But it turns out that $\cos^2(\arctan(x)) = \frac{1}{{1+x^2}}$.  To see this, you can draw a right triangle with vertices at $(0,0), (1, 0)$, and $(1, x)$.  The angle at the origin is $\arctan(x)$, and you can easily compute its cosine. (Try it!)
Or, you can try some algebra. Using the notation from above, we have
$$
\tan(y) = x
$$
$$
\frac{\sin(y)}{\cos(y)} = x
$$
$$
\sin(y) = x\cos(y)
$$
$$
\sin^2(y) = x^2\cos^2(y)
$$
$$
1-\cos^2(y) = x^2 \cos^2(y)
$$
$$
1 = \cos^2(y)(1 + x^2),
$$
$$
\frac{1}{1 + x^2} = \cos^2(y),
$$
as desired.
