why if we pad any signal with enough zeros we can get the same result
Hint:  try computing the linear convolution of $~[1\quad 1\quad  1]~$ and 
$~[0\quad 1 \quad 1]~$ by hand and the circular convolution of the zero-padded
versions
$~[1\quad 1\quad  1 \quad 0\quad 0\quad 0]~$ and $~[0\quad 1 \quad 1\quad 0\quad 0\quad 0]~$ also by hand (or via FFTs if you are desperate enough).