Simplification of $(e^{2x}-1)/(e^x-1)$

Question

Why is $$\frac{e^{2x}-1}{e^x-1}$$ equal to $e^x+1$ ?

Answer 1

Substitute $t = e^x \implies e^{2x} = (e^x)^2 = t^2$, which translates the expression to $\dfrac{t^2 - 1}{t-1} = \dfrac{(t-1)(t+1)}{(t-1)}$. Then what is the result when replacing $t = e^x$?

Answer 2

Note: $(e^x-1)(e^x+1)=(e^x)^2+e^x-e^x-1=e^{2x}−1$
Also note that the denominator $e^x-1\ne0\implies e^x\ne1\implies x\ne0$

$$\frac{e^{2x}−1}{e^x−1}=\frac{(e^x-1)(e^x+1)}{e^x−1}=e^x+1 \text{ where } x\ne0$$