2022/042) Solve for x $(1+ \frac{1}{x})^{x+1} = (1+ \frac{1}{n})^n$

 we have

$(1+ \frac{1}{x})^{x+1} = (\frac{x+1}{x})^{x+1}$

$=(\frac{x}{x+1})^{-(x+1)}$

$=(1- \frac{1}{x+1})^{-(x+1)}$

$= (1+ \frac{1}{-(x+1)})^{-(x+1)}$

comparing with the RHS we get -(x+1) = n  or x = -(n+1)