Expressing $\sum\limits_{k=1}^{n-1}\frac{1}{k(k+1)}$ as $1 - \frac{1}{n}$

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I was reading from INTRODUCTION TO ALGORITHM (THIRD EDITION) By Thomas H. Cormen and came across Telescoping series in the Appendix A (page 1148). And this was the definition:

For any sequence $a_0,a_1,...,a_n$

$\sum\limits_{k=0}^{n-1}{(a_k - a_{k+1}) = a_0 - a_n}$

They then gave an example $\sum\limits_{k=1}^{n-1}\frac{1}{k(k+1)} = \sum\limits_{k=1}^{n-1}(\frac{1}{k} - \frac{1}{k+1})$

= $1 - \frac{1}{n}$

Please can someone help me on how they had $1 - \frac{1}{n}$