Chapter 12 - Section 12.6 - The Binomial Theorem - 12.6 Exercises - Page 887: 57

$(101!)^{100}$

Step 1. Rewrite the two expressions as $a=(100!)^{101}=100\times(100!)^{100}$ and $b=(101!)^{100}=(101\times100!)^{100}=101^{100}(100!)^{100}$ Step 2. Examine the ratio of the two numbers: $\frac{b}{a}=\frac{101^{100}(100!)^{100}}{100\times(100!)^{100}}=\frac{101^{100}}{100}\gt1$ Step 3. Conclusion: $(101!)^{100}$ is larger.