Young Alan Turing and the Arctangent

With the release of the new film "The Imitation Game", I decided to read the biography this excellent film was based on - Alan Turing: The Enigma. In it, the author Andrew Hodges relates the story that the 15-year-old Alan Turing derived the Maclaurin series for the $\arctan$ function, i.e. $$\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \ldots$$ This is trivial using calculus, but it's explicitly stated that young Alan Turing neither knew nor used calculus. How would you derived such a series without calculus?

This is a tricky problem, and I'd suggest first tackling the much easier problem of deriving the Maclaurin series for $\exp(x)$ from the relation $\exp(2x) = \exp(x)\cdot \exp(x)$. This is an underconstrained relation, so you'll need to assume $c_0 = 1, c_1 = 1$.

Getting back to $\arctan$, you could start with the half-angle formula for the tangent: $$\tan(2x) = \frac{2\tan(x)}{1-{\tan}^2(x)}.$$ Now use the Weierstrass-like substitution $x = \arctan(t)$ to get $$\tan(2\arctan(t)) = \frac{2t}{1-t^2}.$$ The right-hand side can be expanded in the usual geometric series fashion to get $$\tan(2\arctan(t)) = 2t\cdot (1+t^2+t^4+\ldots).$$
Finally, take the $\arctan$ of both sides and assume we have the series expansion $\arctan(x) = c_1 x + c_3 x^3 + c_5 x^5 + \ldots$. Note that we may ignore the terms with even powers of $x$ as $\arctan(x)$ is an odd function.

This gives us the setting $$2\arctan(t) = \arctan(2t\cdot (1+t^2+t^4+\ldots))$$ and expanding as a power series $$2(c_1 t + c_3 t^3 + \ldots) = c_1 (2t\cdot (1+t^2+\ldots) + c_3 (2t\cdot (1+t^2+\ldots))^3 + \ldots$$
The next step is to line up powers of $t$ on both sides and set up a system of simultaneous equations. There's some algebra and combinatorics involved, but we end up with the system of equations $$c_{2i+1} = \sum_{j=0}^{i} c_{2j+1}\cdot 2^{2j} \cdot {{i+j} \choose {2j}}.$$ Note that this system is underconstrained due to our functional relationship being satisfied by any multiple of the $\arctan$ function. We'll assume that $c_1 = 1$, but note that this follows from the classical (non-calculus) limit $\lim_{x\to 0} \frac{\sin(x)}{x} = 1$.

The first few relations are $$\begin{align*} c_1 &= c_1 \\ c_3 &= c_1 + 4\cdot c_3 \\ c_5 &= c_1 + 12\cdot c_3 + 16\cdot c_5 \\ c_7 &= c_1 + 24\cdot c_3 + 80\cdot c_5 + 64\cdot c_7 \end{align*}$$
Assuming $c_1 = 1$ as above we quickly calculate $c_3 = -\frac{1}{3}, c_5 = \frac{1}{5}, c_7 = -\frac{1}{7}$, with the pattern being obvious.

That $c_{2i+1} = \frac{(-1)^i}{2i+1}$ can be verified by Wolfram Alpha:

An obvious question is whether or not there's a simple demonstration of this; in particular, one that a young Alan Turing may have found. This I don't know (yet).