已知数列\(\{a_{n}\}\)的首项\(a_{1}=2\)，且\(a_{n+1}=\dfrac{2a_{n}-1}{a_{n}}.\)
\((1)\)证明：数列\(\{\dfrac{1}{a_{n}-1}\}\)为等差数列．
\((2)\)已知\(b_{n}=\lg a_{n}\)，数列\(\{b_{n}\}\)的前\(n\)项和为\(S_{n}\)，若\(S_{m}>2\)，求整数\(m\)的最小值．