已知数列\(\{a _{n} \}\)满足\(a _{1} =1\)，\( \sqrt {a_{n}}- \sqrt {a_{n+1}}= \sqrt {a_{n}\cdot a_{n+1}}(n∈N^{*})\)．
\((\)Ⅰ\()\)求证：数列\(\{ \sqrt { \dfrac {1}{a_{n}}}\}\)为等差数列，并求\(a _{n}\)；
\((\)Ⅱ\()\)设\(b_{n}= \sqrt {1+2a_{n+1}}\)，数列\(\{b _{n} \}\)的前\(n\)项和为\(S _{n}\)，求证：\(S_{n} < n+1- \dfrac {1}{n+1}\)．