Induction proof for $ \sum _ { i = 1 } ^ n \frac 1 { i \cdot ( i + 1 ) } = 1 - \frac 1 { n + 1 } $

Question

Prove that for any positive integer $ n $ applies: $$ \frac 1 { 1 \cdot 2 } + \frac 1 { 2 \cdot 3 } + \frac 1 { 3 \cdot 4 } + \dots + \frac 1 { n \cdot ( n + 1 ) } = \sum _ { i = 1 } ^ n \frac 1 { i \cdot ( i + 1 ) } = 1 - \frac 1 { n + 1 } \text . $$

I have done this so far:

It applies for $ n = 1 $ because $ 1 - \frac 1 { n + 1 } = 1 - \frac 1 { 1 + 1 } = 1 - \frac 1 2 = \frac 1 2 $ but what about $ n = j + 1 $: $$ \sum _ { i = 1 } ^ j \frac 1 { i \cdot ( i + 1 ) } = 1 - \frac 1 { j + 1 } $$ I have substituted $ j $ for $ n $ and now consider the expression for $ j + 1 $, $ \sum _ { i = 1 } ^ { j + 1 } \frac 1 { i \cdot ( i + 1 ) } $: \begin{align*} \sum _ { i = 1 } ^ { j + 1 } \frac 1 { i \cdot ( i + 1 ) } & = \sum _ { i = 1 } ^ j \frac 1 { i \cdot ( i + 1 ) } + \frac 1 { ( j + 1 ) \cdot \bigl ( ( j + 1 ) + 1 \bigr ) } \\ & = 1 - \frac 1 { j + 1 } + \frac 1 { ( j + 1 ) \cdot \bigl ( ( j + 1 ) + 1 \bigr ) } \\ & = 1 - \frac 1 { j + 1 } + \frac 1 { ( j + 1 ) \cdot ( j + 2 ) } \\ & = 1 - \frac { 1 \cdot \bigl ( ( j + 1 ) \cdot ( j + 2 ) \bigr ) } { ( j + 1 ) \cdot \bigl ( ( j + 1 ) \cdot ( j + 2 ) \bigr ) } + \frac { 1 \cdot ( j + 1 ) } { ( j + 1 ) \cdot \bigl ( ( j + 1 ) \cdot ( j + 2 ) \bigr ) } \\ & = 1 - \frac { ( j + 1 ) \cdot ( j + 2 ) + j + 1 } { ( j + 1 ) \cdot \bigl ( ( j + 1 ) \cdot ( j + 2 ) \bigr ) } \\ & = 1 - \frac { ( j + 1 ) \cdot ( j + 2 ) + j + 1 } { ( j + 1 ) ^ 2 \cdot ( j + 2 ) } \\ & = 1 - \frac { j + 1 } { ( j + 1 ) ^ 2 } \cdot \left ( \frac { j + 2 } { j + 1 } + \frac { j + 1 } { j + 2 } \right ) \\ & = 1 - \frac 1 { j + 1 } \cdot \left ( 1 + \frac { j + 1 } { j + 2 } \right ) \\ & = 1 - \frac 1 { j + 1 } + \frac 1 { j + 1 } \cdot \frac { j + 1 } { j + 2 } \\ & = 1 - \left ( \frac 1 { j + 1 } + \frac { j + 1 } { ( j + 1 ) \cdot ( j + 2 ) } \right ) \\ & = 1 - \frac 1 { j + 1 } + \frac 1 { j + 2 } \end{align*}