Order in the Math Room

Firstly, we need to prove that $(n+1)!^{n}$ > $(n)!^{n+1}$ . Now $\frac{(n!)^{n+1}}{(n+1)!^{n}}$ = $\frac{(n!).(n!)^n}{(n!)^n . (n+1)^n}$ = $\frac{n!}{(n+1)^n}$ < 1, as required. Now using this fact: $(70)!^{69}$ > $(69)!^{70}$ >...> $(6!)^{133}$ > $134^{134}$ , as $\frac{(6!)^{133}}{134^{134}}$ > $5^{133}$ $\times$ $\frac{1}{134}$ > 1. Now we need to deal with $69^{69^{69}}$ , which is where we have to be careful. Let ${a}$ = $69^{69}$ such that the expression becomes $69^{a}$ , but not $a^{69}$ , however the 2nd expression will come in handy. It is clear that $69^{a}$ > $a^{69}$ for large values of ${a}$ , which is what we have. Now ${a}$ > 70!, so $69^{a}$ > $a^{69}$ > $(70)!^{69}$ . This produces the required result.