Abnormally Normal

Implicit differentiation of $k^x+y^k+x^x=x^k+k^y+y^y$ gives $(k^x)(\ln{k}) + (k)(y^{k-1})\left(\frac{dy}{dx}\right) + (x^x)(\ln{x}+1) = (k)(x^{k-1}) + (k^y)(\ln{k})\left(\frac{dy}{dx}\right) + (y^y)(\ln{y}+1)\left(\frac{dy}{dx}\right)$ $(k)(y^{k-1})\left(\frac{dy}{dx}\right) - (k^y)(\ln{k})\left(\frac{dy}{dx}\right) - (y^y)(\ln{y}+1)\left(\frac{dy}{dx}\right) = (k)(x^{k-1}) - (k^x)(\ln{k}) - (x^x)(\ln{x}+1)$ $\left(\frac{dy}{dx}\right) \left((k)(y^{k-1}) - (k^y)(\ln{k}) - (y^y)(\ln{y}+1)\right) = (k)(x^{k-1}) - (k^x)(\ln{k}) - (x^x)(\ln{x}+1)$ $\frac{dy}{dx} = \frac{(k)(x^{k-1}) - (k^x)(\ln{k}) - (x^x)(\ln{x}+1)}{(k)(y^{k-1}) - (k^y)(\ln{k}) - (y^y)(\ln{y}+1)}$ For constant $k$ , at every point $(t,t)$ where $t \in \mathbb{Z^+}$ , $t \geq k$ , $\frac{dy}{dx}=1$ , hence $G_{k,t}=-1$ . Therefore $\prod_{k=2}^{2016} \left( \prod_{t=k}^{k+2015} G_{k,t} \right) = 1$