The significance of $\pi=\frac{\ln(-1)}i$ This is one way of defining $\pi$ but what does this definition tell us? This is undefined but $\pi$ isn't undefined.
$$e^{i\pi}+1=0$$
$$\pi=\frac{\ln(-1)}i$$
 A: $\ln(-1)$ isn't undefined, contrary to what is commonly taught in precalculus.
If we allow complex numbers $z = x+iy$ as input, then we can define the so-called principal value of the complex logarithm, denoted $\operatorname{Log}(z)$, as $$\operatorname{Log}(z) = \ln |z| + i\theta.$$  Here, $|z| = \sqrt{x^2+y^2}$ is the modulus of $z$, and $\theta \in (-\pi, \pi]$ is called the argument of $z$, which is, very loosely speaking, the angle that $z$ makes with the positive $x$-axis.  Note that we can define the complex logarithm slightly differently so that we don't have the principal value, but then we get into the concept of branch cuts and I think that is beyond the scope of necessary discussion here.
If we interpret $-1$ as the complex number $-1 + 0i$, then we have $$\operatorname{Log}(-1+0i) = \ln\left(\sqrt{(-1)^2 + 0^2}\right) + i\pi = \ln|1| + i\pi = i\pi.$$
This is how we can have $\ln(-1) = i\pi$, and therefore $\pi = \dfrac{\ln(-1)}{i}$.
In case it's not clear, the reason that $\theta = \pi$ for $-1 + 0i$ is because when we plot $-1+0i$ in the $xy$-plane, we plot it at the point $(-1,0)$, a full $\pi$ radians away from the positive $x$-axis.  And since $-\pi < \theta \le \pi$, then we take $\theta = \pi$ (and not, say, $-\pi$, or $3\pi$, etc.).
Regarding its significance.. Apart from being interesting and perhaps surprising at first, this isn't really a significant fact regarding $\pi$ in the sense that it has wide applications or gives us insight into anything regarding the digits of $\pi$.  Not as far as I know, anyway.
A: Not a bad formula but unfortunately it uses a transcendental function to define $\pi$, which is not that useful.  It is perhaps more useful to provide a formula for $\pi$ that only uses roots; see this article (section 3.4).
A: One way to view Euler's identity as saying that multiplication by $e^{i\theta}$ for real $\theta$ corresponds to counterclockwise rotation in the complex plane by $\theta$ radians. We can see this in the following manner. Let us take as given that $\lim_{n \to \infty} (1+i \theta/n)^n = e^{i \theta}$ (either by defining things this way or proving that whatever definition we use has this property). 
Notice that multiplication by $1+i\theta/n$, just from the definition of complex multiplication, is the same as multiplication by the matrix $R(\theta/n)$ where $R(x)$ is defined as 
$$\begin{bmatrix} 1 & -x \\ x & 1 \end{bmatrix}.$$ 
So multiplication by $(1+i\theta/n)^n$ is multiplication by $R(\theta/n)^n$.
Suppose now $\theta/n \in [-1,1]$ (always true for large enough $n$). From geometry, we know that $R(\theta/n)$ represents a rotation by $\arcsin(\theta/n)$ radians and a dilation by $(1+(\theta/n)^2)^{1/2}$. Thus the problem reduces to showing that $\lim_{n \to \infty} (1+(\theta/n)^2)^{1/2} = 1$ and $\lim_{n \to \infty} n \arcsin(\theta/n) = \theta$. These two calculations are routine and also have relatively simple geometric meaning.
