I am supposed to tell which one of $(101!)^{100}$ and $(100!)^{101}$ is larger. I am trying to use the behavior of the function $f(x)=x^{1/x}$ as is a standard technique to dealing with questions of this sort. Here is what I have so far.
$$\begin{aligned}(101!)^{100!}&\lt (100!)^{101!}\\ (101!)^{100} &\lt (100!)^{101\times 100}\end{aligned}$$
Any ideas on how to proceed. Thanks.
Well
$$(101!)^{100}\cdot (101!)=\color{green}{(101!)^{101}}$$ while: $$(100!)^{101}\cdot (101)^{101}=\color{green}{(101!)^{101}}$$
the one you have to multiply by the larger number is smaller
Simplify: \begin{align} \frac{(101!)^{100}}{(100!)^{101}}&=\frac{(100!\times 101)^{100}}{(100!)^{100}\times(100!)}\\ &=\frac{(100!)^{100}\times 101^{100}}{(100!)^{100}\times(100!)}\\ &=\frac{101^{100}}{100!} \end{align} Now, $$\frac{101^{100}}{100!}=\frac{101\times101\times101\times\dots_{100\mathrm{~times}}}{100\times99\times98\times\dots2\times1}=\frac{101}{100}\times\frac{101}{99}\times\frac{101}{98}\times\dots\frac{101}{2}\times\frac{101}{1}>1.$$ You get
$$\frac{(101!)^{100}}{(100!)^{101}}=\frac{101^{100}}{100!}>1\Rightarrow(101!)^{100}>(100!)^{101}$$
Consider $$A_n=\big[(n+1)!\big]^n \qquad \text{and}\qquad B_n=\big[n!\big]^{n+1}$$ $$R_n=\frac{A_n}{B_n}=\frac{\big[(n+1)!\big]^n } {\big[n!\big]^{n+1} }$$ Take logarithms $$\log(R_n)=n \log ((n+1)!)-(n+1) \log (n!)$$ Use Stirling approximation and continue with Taylor expansions $$\log(R_n)=n+\left(1-\frac{1}{2} \log (2 \pi n)\right)-\frac{7}{12 n}+O\left(\frac{1}{n^2}\right)$$ $$R_n\sim\frac{e^{n+1}}{\sqrt{2 \pi n}} $$
For $n=100$, the above would give $R_{100}\sim 2.92\times 10^{42}$ while the exact value would be $2.90\times 10^{42}$
