How to evaluate the following Product?

Question

so I have the following product to evaluate : $$P_{n}=\prod_{k=0}^{n-1} u_{k}$$ Where : $u_k = e^{w_n}$ All what I know about $u_k$ is that $$ u_{n}=\frac{n+2}{n+1} $$ So we'll have the following : $$ \begin{aligned} u_{n}=e^{w_{n}} \Leftrightarrow w_{n} &=\ln \left(u_{n}\right) \\ \Leftrightarrow w_{n} &=\ln \left(\frac{n+2}{n+1}\right) \\ &=\ln (n+2)-\ln (n+1) \end{aligned} $$ And I transformed the product to the following : $$ \begin{aligned} P_{n} &=\prod_{k=0}^{n-1} u_{k} \\ &=e^{w_{0}} \times e^{w_{1}} \times \cdots \times e^{w_{n-1}} \\ &=e^{w_{0}+w_{1}+\cdots+w_{n-1}} \\ \ln \left(P_{n}\right) &=w_{0}+w_{1} \ldots + w_{n-1} \\ &=(\ln 2-\ln 1)+(\ln 3-\ln 2)+\ldots+( \ln (n+1)-\ln (n)) \end{aligned} $$ And yeah from here I got kind confused because of the elimination of terms, can someone guide me ? And if someone is interested it's an exercise from a Moroccan exam, thanks in advance.

Answer 1

$\ln (2)$ cancels with the $-\ln (2)$ and $\ln (3)$ cancels with the $-\ln (3)$ and so on leaving $\ln (n+1) -\ln (1)$

edit: unless you have to use $e$ and $ln$ then

$$P_n = \prod_{k=0}^{n-1}u_n=\prod_{k=0}^{n-1}\frac{k+2}{k+1}=\prod_{k=0}^{n-1}\frac{2}{1}\frac{3}{2}\frac{4}{3}.....\frac{n}{n-1}\frac{n+1}{n}$$ and cancelling diagonly i.e. 2 with $\frac{1}{2}$ and so give $P_n = n+1$

Is that clear?