Hint: Compare $(8!)^9=(8!)(8!)^8$ and $(9!)^8=9^8(8!)^8$.
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Define the function $f(x) = x^{\frac{1}{x}}$ and calculate its derivative. You should get $f'(x)=-x^{\frac{1}{x}-2}(ln(x)-1)$. You see that that $\forall x > e: f'(x) < 0$ since $ln(e)=1$. Thus, the function $f(x)$ is strictly decreasing function for $x>e \approx 2.718$. Hence, $f(8) > f(9)$.
Hint: Compare $(8!)^9=(8!)(8!)^8$ and $(9!)^8=9^8(8!)^8$.