Prove that $\sqrt{-1}=\frac{i}{\pi}$ It's highly unlikely but suppose human beings first considered the problem of $\ln{(-1)}$ before $\sqrt{-1}$ and gave $\ln{(-1)}$ the symbol $i$.
Now, can we work our way backwards to show that 
$$\sqrt{-1}=\frac{i}{\pi}$$
I don't think that the Taylor series proof of Euler's identity can work backwards.
 A: The problem with your question as it stands is that $i := \ln(-1)$ is not a proper definition.
What we are doing when we are introducing the complex numbers is that we are introducing a new "symbol" $i$ and then let $\Bbb{C} = \{ a + bi \,:\, a,b\in \Bbb{R}\}$. If you don't like the idea of a "symbol" $i$, then let $\Bbb{C} = \{(a,b) \,:\, a,b\in \Bbb{R}\}$ and simply interpret $a+bi$ as another way of writing $(a,b)$.
Then we are defining the arithmetic operations $+,\cdot$ on $\Bbb{C}$ in such a way that $i^2 = -1$. But we still have defined $+,\cdot$ on all of $\Bbb{C}$.
In your case, you just say "let $i = \ln (-1)$". This gives us no further idea about the properties of $i$. In particular, we don't know how to calculate with it. Also, "$i = \ln(-1)$" is not really meaningful. I guess this to mean $\exp(i) = -1$. But we then still have to define $\exp$ on some set which contains $i$ and $i/\pi$ (since you are considering $i/\pi$, you also have to explain what it means to divide the "symbol" $i$ that you introduced by $\pi$).

Finally, there is another problem. Let us assume that your $i := \ln(-1)$ is a complex number. I will write $j$ for the "real" imaginary unit, since you insist on calling your new symbol $i$.
Then, $\exp(i) = -1$. But we also have $\exp(i + 2\pi j) = -1$. So if the only property you have on $i$ is that $\exp(i)=-1$, then you will not be able to show $i / \pi = j$, since the property $\exp(i)=-1$ does not determine $i$ uniquely.
A: The given definition is a little problematic, because of
$$\ln(-1)=-\ln\left(\frac1{-1}\right)=-\ln(-1),$$ hence $$i=-i.$$
This needs to be somehow fixed.
