Proving $e^{i\pi} = -1$ without proving $e^{ix} = \cos x + i\sin x$ Just as the title asks: I'm looking for a proof of Euler's identity without first proving the general case of $e^{ix} = \cos x + i\sin x$.
I outlined in a previous question a proof that $$\lim_{n\to\infty}\left(1+\frac{ix}{n}\right)^n = \cos x + i\sin x$$

That outline (about halfway through the answer) can be altered just to prove it for $x=\pi$ pretty directly. Basically:

$$\frac{\cos(\pi/n)+i\sin(\pi/n)}{1+i\pi/n}=1+g(n)$$

where $ng(n)\to 0$, so we can prove that:$$\left(\frac{\cos(\pi/n)+i\sin(\pi/n)}{1+i\pi/n}\right)^n\to 1$$

You can prove that $ng(n)\to 0$ geometrically.  You essentially only need to prove that:

$$\left|\,\cos(\pi/n) - 1\right| = O(1/n^2)$$ and $$\left|\,\sin(\pi/n)-\pi/n\right| = O(1/n^2)$$

The first is easy to prove. The distance from $(\cos x,\sin x)$ to $(1,0)$ is $\sqrt{2-2\cos x}$ and $x$ is the length of the arc of the circle between those two points.  So $$0\leq 1-\cos\pi/n\leq \frac{\pi^2}{2}\frac{1}{n^2}$$

The second requires more care. We can easily see $0\leq \sin\pi/n \leq \pi/n$ since $\sin \pi/n$ is the distance of $(\cos\pi/n,\sin\pi/n)$ to the real line, and $\pi/n$ is the length of a longer path to the real line. 

Now, take the points $P=(1,0)$ and $Q=(\cos 2\pi/n,\sin 2\pi/n)$. Draw the tangents to the unit circles at these points and find their intersection, $R$. Then show that the length of
the path $PRQ$ is $2\tan \pi/n$. This is a path "outside" the circle, so it must be greater than the path of length $2\pi/n$ along the circle between $P$ and $Q$, so we get inequality: $$\tan\pi/n \geq \pi/n$$ This gives us the inequality:

$$\frac{\pi}{n}\cos \pi/n \leq \sin\pi/n \leq\frac{\pi}n$$

Which means that $$|\pi/n-\sin\pi/n|\leq \frac{\pi}{n}(1-\cos\pi/n) = O(1/n^3)$$
Just take the log of both sides:

$$\ln e^{i\pi}=\ln(-1)\Rightarrow i\pi\ln e=i\pi\Rightarrow i\pi=i\pi$$