Is $2\pi /2\pi$ considered an indeterminate ratio for imaginary exponentiation? It seems that I can force any expression of the form $a^{bi}$ to become 1 by changing its base to the natural number and raising it to the power of one in the form of $\frac{2\pi}{2\pi}$ so that it instead reads as $(e^{2πi})^\frac{blna}{2π}$ or $1^c$, $a,b,c\in\mathbb{R}$.
Does this problem arise from the periodic similarity between $0$ and $2\pi$ in the trigonometric representation, meaning that in this context $\frac{2\pi}{2\pi}$ is akin to the indeterminate ratio $\frac{0}{0}$?  This kind of mistake can be put into any false proof, but I have never seen it come up without zeroes or infinities. If not is there some restrictions on the properties I've used that prevents them from behaving in the way I understand way for imaginary exponents?  
 A: When using powers of complex numbers you will have to be careful, because in general they are not uniquely defined.
This leads to the power laws are not valid in a unchanged form, that is $(z^u)^v$ is only equal to $z^{uv}$ in some restricted sense.
The definition of complex powers is reduced to the definition of the (natural) logarithm (which in turn is not uniquely determined). We have
$$a^b = e^{b \ln a}$$
so we have
$$z^{uv} = e^{uv \ln z}$$
while
$$(z^u)^v = e^{v \ln z^u} = e^{v \ln e^{u \ln z}}$$
now it would be tempting to just use that $e^{\ln w} = w$, but that's only true in the real case, in the complex case we would have to take into account the multi-valuedness of $\ln$ so $e^{\ln w} = w + 2n\pi i$ which gives:
$$(z^u)^v = e^{v \ln e^{u \ln z}} = e^{v (u\ln z + 2n\pi i)} = e^{uv\ln z + v2n\pi i} = z^{uv}e^{2vn\pi i}$$
So let's see where your approach gets you:
$$(e^{2\pi i})^{b\ln a/2\pi} = e^{ib\ln a}e^{inb\ln a} = e^{i(n+1)b\ln a}$$
Now it may seem to clear, if $n=-1$ you actually get this to be $1$, but that one need not be $a^{bi}$, it won't if $b\ne 0$.
