I need a proof of the following result to calculate a Nash equilibrium in the Showcase Showdown game.

For all $n>1$, the system of equations $$\left\{ \begin{aligned} (1+e^{x}(-1+x))^{n-2}&=\dfrac{e^{y}\,\left( -1+{\left( 1+e^{x}\,\left( -1+y\right) \right) }^{n}\right) +e^{x}\,n}{e^{x}\,\left( 1+e^{y }\,\left( -1+y\right) \right) \,n\,\left( 1+e^{x}\,\left( -2+n+x\right) \right) }\\ {(1+e^{x}(-1+x))}^{n-1}&=\dfrac{-1+{\left( 1+e^{x}\,\left( -1+y\right) \right) }^{n}+e^{x}\,{\left( 1+e^{x}\,\left( -1+y\right) \right) }^{n-1}\,n}{e^{x}\,\left( 1+e^{y}\,\left( -1+y\right) \right) \,n} \end{aligned} \right.$$ has a unique solution $(\epsilon_n,\delta_n)\in[0,1]\times[0,1]$ with $\delta_n>\epsilon_n$.