summation - What is the sum of reciprocals of first $n$ Euclid numbers

We know about Euclid Number . I want to know the sum of reciprocals of 1st $n$ Euclid Number ?
In that book I have been told to find the value of following series :

$$\dfrac{1}{e_1}+\dfrac{1}{e_2}+\cdots+\dfrac{1}{e_n} = ?$$

In fact, I can not understand the following calculations:
$$\dfrac{1}{e_1}+\dfrac{1}{e_2}+\cdots+\dfrac{1}{e_n} = 1 - \dfrac{1}{e_n(e_{n}-1)} = 1-\dfrac{1}{e_{n+1}-1}$$

Can you please help me to find how this sum is done? This math is taken from "Concrete Math" Of Ronald L. Graham, Donald E. Knuth, and Oren Patashnik.

The definition in this book for Euclid numbers is non-standard: $e_1=2$ and $e_{n+1}=e_1\dots e_n +1$.

Answer

Given $e_1=2$ and $e_{n+1}=e_1\dots e_n+1$, then:

$$\frac{1}{e_1}+...+\frac 1 {e_n} = 1-\frac 1 {e_1\dots e_n} = 1-\frac 1 {e_{n+1}-1}$$

This can easily be proved by induction. Just compute:

$$\frac{1}{e_1}+...+\frac 1 {e_n} + \frac{1}{e_{n+1}} = 1-\frac 1 {e_{n+1}-1} + \frac{1}{e_{n+1}}$$

Just multiply it out and used that $e_{n+2}-1 = e_{n+1}(e_{n+1}-1)$.

You get the middle term of your question by noting that $$e_1\dots e_n = (e_1\dots e_{n-1})e_n = (e_n-1)e_n$$