Solution 1:

How about $$ \sum\limits_{n = 1}^\infty {\frac{{I_n }}{{n!}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{n!}}\int_0^1 {e^{ - y} y^n dy} } = \int_0^1 {e^{ - y} \sum\limits_{n = 1}^\infty {\frac{{y^n }}{{n!}}} dy} = \int_0^1 {e^{ - y} (e^y - 1)dy} = e^{ - 1} \,? $$