1. The problem asks about the logarithm of infinity factorial, that is \( \log(\infty!) \). 2. First, note that \( \infty! \) is not defined because factorial is defined only for non-negative integers and infinity is not a number. 3. However, in analysis and asymptotics, we consider the growth of \( n! \) as \( n \to \infty \). 4. Using Stirling's approximation: $$ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n $$ 5. Taking the logarithm of both sides for large \( n \): $$ \log(n!) \approx \log\left(\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \right) = \log(\sqrt{2 \pi n}) + \log\left(\left(\frac{n}{e}\right)^n\right) $$ 6. Simplify the right hand side: $$ = \frac{1}{2}\log(2\pi n) + n \log(n) - n \log(e) $$ 7. Since \( \log(e) = 1 \), this becomes: $$ \frac{1}{2} \log(2 \pi n) + n \log(n) - n $$ 8. For very large \( n \), the dominant term is \( n \log(n) \) which grows without bound. 9. Therefore, we say: $$ \lim_{n \to \infty} \log(n!) = \infty $$ 10. In summary, \( \log(\infty!) \) is not rigorously defined but the logarithm of factorial grows to infinity as \( n \to \infty \).