208. Comparison of radicals. Which of the two quantities is greater: $\sqrt[8]{8!}$ or $\sqrt[9]{9!} ?$
208. In general, $n!<(n+1)^{n}$, since each of the $n$ factors on the left side is less than $(n+1)$. Therefore,

$$
(n!)^{n} n!<(n!)^{n}(n+1)^{n} \quad \text { or } \quad(n!)^{n+1}<[(n+1)!]^{n} .
$$

Taking the $n(n+1)$-th root of both sides of this inequality, we get

$$
(n!)^{\frac{1}{n}}<[(n+1)!]^{\frac{1}{(n+1)}}
$$

In our specific case, we could reason as follows. Suppose that $(8! )^{\frac{1}{8}} \geqslant(9!)^{\frac{1}{9}}$. Raising both sides of this inequality to the 72nd power, we would arrive at the inequality $(8!)^{9} \geqslant(9!)^{8}$, the falsity of which becomes evident if we divide both sides by $(8!)^{8}$. (Indeed, we would then get $8! \geqslant 9^{8}$, which is false.) Thus, $\sqrt[8]{8!}<\sqrt[9]{9!}$.