已知数列\(\{a _{n} \}\)前\(n\)项和为\(S _{n}\)，且满足\(a_{1}=1,a_{n}=-S_{n}S_{n-1}(n\geqslant 2,n∈N^{*})\)．
\((\)Ⅰ\()\)求数列\(\{a _{n} \}\)的通项公式；
\((\)Ⅱ\()\)记\(T _{n}\)为\(\{a _{n+1} S _{n} \}\)的前\(n\)项和，\(n∈N ^{*}\)，证明：\(T_{n}\geqslant - \dfrac {3}{4}+ \dfrac {1}{2n(n+1)}\)．