1. **Problem statement:** Find the value of the infinite series $\sum_{k=1}^\infty \frac{1}{k!}$. 2. **Recall the exponential series:** The Maclaurin series expansion for $e^x$ is given by $$e^x = \sum_{k=0}^\infty \frac{x^k}{k!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$ 3. **Apply the formula for $x=1$:** $$e^1 = e = \sum_{k=0}^\infty \frac{1^k}{k!} = \sum_{k=0}^\infty \frac{1}{k!} = 1 + \sum_{k=1}^\infty \frac{1}{k!}$$ 4. **Isolate the series:** $$\sum_{k=1}^\infty \frac{1}{k!} = e - 1$$ 5. **Interpretation:** The series sums all reciprocals of factorials starting from $k=1$, which converges to $e - 1$. **Final answer:** $$\sum_{k=1}^\infty \frac{1}{k!} = e - 1$$