The A3 problem in the 2015 Putnam competition is this:
$$\log_{2}\left(\prod_{a=1}^{2015}\prod_{b=1}^{2015}\left(1+e^{\frac{2\pi iab}{2015}}\right)\right)$$
My solution:
$$e^{2\pi iab}=(((e^{\pi i})^2)^{\frac{ab}{2015}}$$
$$=((-1)^{2})^{\frac{ab}{2015}}=(1)^{\frac{ab}{2015}}=1$$
The expression becomes
$$\log_{2}\left(\prod_{a=1}^{2015}\prod_{b=1}^{2015}2\right)$$
$$=\log_{2}(\prod_{a=1}^{2015}2^{2015})=\log_{2}(2^{2015^2})=2015^2=4060225$$
The answers states the answer is 13725. I feel like my reasoning is sound. Why is the answer not 4060225? My theory is that there are 2015 2015th roots of one and their solution used a different one, but both answers are correct.
Your error is in asserting that $1^{1/n} = 1$ for all positive integer $n$. While this is true if we are considering only the positive real root of a number, it is not true for complex numbers. Recall that the $n^{\rm th}$ roots of unity $$1, \zeta, \zeta^2, \ldots, \zeta^{n-1}, \\ \zeta = e^{2 \pi i/n} = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}$$ are solutions to the polynomial $$z^n - 1 = 0.$$ So you cannot write something like $$e^{2 \pi i ab/2015} = (e^{2 \pi i})^{ab/2015} = 1^{ab/2015} = 1$$ because even the trivial choice $a = b = 1$ gives $$e^{2\pi i/2015} = \cos \frac{2\pi}{2015} + i \sin \frac{2\pi}{2015} \ne 1.$$