Power of 2000

Let $y=-x-1$ and we plug that into

${ \left( 1+\dfrac { 1 }{ y } \right) }^{ y+1 }$

${ \left( 1-\dfrac { 1 }{ x+1 } \right) }^{- x }$

${ \left( \dfrac { x }{ x+1 } \right) }^{ -x }$

${ \left( \dfrac { x+1 }{ x } \right) }^{ x }$

${ \left( 1+\dfrac { 1 }{ x } \right) }^{ x }$

Hmm….

We will manipulate the LHS until it has the form of the RHS $LHS = \left ( 1+ \frac{1}{x} \right )^{x+1} = \left(\frac{x+1}{x}\right)^{x+1}$ while $RHS = \left (1 + \frac{1}{2000}\right )^{2000} = \left(\frac{2001}{2000}\right)^{2000}$

LHS is not yet in the RHS form because the denominator must equal the power, and each must be one less than the numerator. So we manipulate LHS.

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

We manipulate the fraction in the LHS further, multiplying numerator and denominator by -1 so that the denominator and the power are equal, as it is in the RHS

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

Finally we can compare with the RHS and get $-x-1 = 2000$

$\Rightarrow x = -2001$

(1+1/x)^(x+1)=(1+1/2000)^2000

>>> ((x+1)/x)^(x+1)=(2001/2000)^2000 Now you notice that if x=2000 then the numerator would be the same as the RHS numerator but the power won't be, but on taking -2001 the equation will hold ((-2001+1)/-2001)^-2001+1 =( -2000/-2001)^-2000 = ((2000/2001)^(-1))^2000 =(2001/2000)^2000 =RHS