Recall the series expansion of the exponential function,
[tex]\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!}[/tex]
By comparison, the given sum is [tex]e^x[/tex] evaluated at x = ln(2), so
[tex]\displaystyle 1 + \ln(2) + \frac{\ln^2(2)}{2!} + \cdots = \sum_{n=0}^\infty \frac{\ln^n(2)}{n!} = e^{\ln(2)} = \boxed{2}[/tex]
