1. The problem involves evaluating the infinite sum expression involving complex exponentials and factorials, specifically expressions of the form $$\sum_{k=0}^\infty \frac{1}{k!}$$ raised to powers involving $$i\pi$$. 2. Recall the exponential series expansion: $$e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$ 3. Using this, we identify: $$\sum_{k=0}^\infty \frac{1}{k!} = e^1 = e$$ 4. Therefore, the expression $$\left(\sum_{k=0}^\infty \frac{1}{k!}\right)^{i\pi} = e^{i\pi}$$. 5. Using Euler's formula: $$e^{i\pi} = \cos(\pi) + i\sin(\pi) = -1 + 0 = -1$$ 6. Hence, the expression simplifies to $$-1$$. 7. Any powers or products involving this term can be simplified accordingly, for example: $$\left(e^{i\pi}\right)^2 = e^{2i\pi} = 1$$ 8. The complicated nested sums and products involving $$\sum_{k=0}^\infty \frac{1}{k!}$$ raised to powers of $$i\pi$$ reduce to powers of $$-1$$ or 1, depending on the exponent. 9. Therefore, the entire expression simplifies to a finite value based on powers of $$-1$$ and 1. 10. Final answer: $$\boxed{-1}$$ This is a fundamental result from Euler's identity and the exponential series expansion.