Use induction to prove that if 
for all 
, then
![\[ \prod_{k=1}^n \frac{a_k}{a_{k-1}} = \frac{a_n}{a_0}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-78568722bdd5abfdae7828291ad40d28_l3.png "Rendered by QuickLaTeX.com")
This is the telescoping property for products.
Proof. For the case
we have,
![\[ \prod_{k=1}^1 \frac{a_k}{a_{k-1}} = \frac{a_1}{a_0}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-789d2e9c8e6b780e2a9e653d0873e672_l3.png "Rendered by QuickLaTeX.com")
Thus, the property holds for the case . Assume then that it holds for some 
. Then,
![\[ \prod_{k=1}^{m+1} \frac{a_k}{a_{k-1}} = \left( \frac{a_{m+1}}{a_m} \right) \cdot \prod_{k=1}^m \frac{a_k}{a_{k-1}} = \left( \frac{a_{m+1}}{a_m} \right) \left( \frac{a_m}{a_0} \right) = \frac{a_{m+1}}{a_0}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a91f8c8d7b97e514aa3aee995ca8e324_l3.png "Rendered by QuickLaTeX.com")
Hence the property is true for 
; and thus, for all 