$$\left( 1 + \frac{1}{2} \right) \left( 1 + \frac{1}{3} \right) \left( 1 + \frac{1}{4} \right) ... \left( 1 + \frac{1}{99} \right) $$

Of course I could do it factor for factor but there has got to be a more efficient way. Does anyone know it?