Answer
See below
Work Step by Step
$$(100!)^{101} = (100!)^{100} *100!$$
$$(101!)^{100} = {(100!)}^{100}*101^{100}$$
$100!\lt (101!)^{100}$ because
$100*99*98*97*...$ has 100 terms all less than 101.
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305115546
ISBN 13:
978-1-30511-554-5
Chapter 8, Sequences and Series - Section 8.6 - The Binomial Theorem - 8.6 Exercises - Page 637: 57
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