How I can show that $\prod_{i=1}^{n}\frac{3^i(x+1)-2^i}{3^i(x+1)-3\cdot2^{i-1}}=\frac{1}{x}-\frac{1}{x}\left(\frac{2}{3}\right)^n+1$? How I can show that the following equality is true: 
$$\prod_{i=1}^{n}\dfrac{3^i(x+1)-2^i}{3^i(x+1)-3\cdot2^{i-1}}=\dfrac{1}{x}-\dfrac{1}{x}\left(\dfrac{2}{3}\right)^n+1\,?$$

Here we have a telescoping product.

We obtain for $n\geq 1$:
  \begin{align*}
\color{blue}{\prod_{j=1}^n\frac{3^j(x+1)-2^j}{3^j(x+1)-3\cdot 2^{j-1}}}
&=\frac{1}{3^n}\prod_{j=1}^n\frac{3^j(x+1)-2^j}{3^{j-1}(x+1)-2^{j-1}}\tag{1}\\
&=\frac{1}{3^n}\,\frac{\prod_{j=1}^n \left(3^j(x+1)-2^j\right)}{\prod_{j=1}^{n}\left(3^{j-1}(x+1)-2^{j-1}\right)}\\
&=\frac{1}{3^n}\,\frac{\prod_{j=1}^n \left(3^j(x+1)-2^j\right)}{\prod_{j=0}^{n-1}\left(3^{j}(x+1)-2^{j}\right)}\tag{1}\\
&=\frac{1}{3^n}\,\frac{3^n(x+1)-2^n}{(x+1)-1}\tag{2}\\
&=\frac{3^nx+3^n-2^n}{3^nx}\\
&\,\,\color{blue}{=1+\frac{1}{x}-\left(\frac{2}{3}\right)^n\frac{1}{x}}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we factor out $\frac{1}{3^n}$.

*In (2) we shift the index in the product of the denominator by one to start with $j=0$.

*In (3) we use the telescopic property of the product and cancel equal factors in numerator and denominator.
