title: Is nlogn same as n^2logn^2
body: <p>When we calculate Big O, is <em>n</em>log<em>n</em> the same as <em>n</em>^2log<em>n</em>^2 ?(basically when <em>n</em> is replaced with <em>n</em> square)</p>
  <p>I was discussing this today with a colleague who mentioned the <em>n</em>log<em>n</em> complexity of an algorithm is same when <em>n</em> becomes <em>n</em>^2.</p>
  <p>That does not sound correct. Looking for expert opinions.</p>