Show $7!^{1/7} < 8!^{1/8}$

Question

Show $7!^{1/7} < 8!^{1/8}$

So I know that the first step is to remove the radicals. So would I raise both sides to the power of 8 to get $({7!}^{1/7})^8 < 8!$. I am not sure where to go from here, I am sure there is some trick I don't know to solve this.

Answer 1

Hint: Note that $$7!<8^7$$ then

$$7!<8^7\implies(7!)^\frac{1}{7}<8\implies7!\cdot(7!)^\frac{1}{7}<7!\cdot8\implies(7!)^\frac{8}{7}<8!\implies(7!)^\frac{1}{7}<(8!)^\frac{1}{8}$$

Answer 2

It is claimed that $(7!)^8 < (8!)^7$. Indeed $$(8!)^7=(8\cdot 7!)^7=8^7\cdot (7!)^7 > 8!\cdot (7!)^7>(7!)^8.$$

Answer 3

You could raise both sides to the power $56$ so that you are comparing $7!^8$ and $8!^7$. Then $$8!^7=8^7\cdot 7!^7>7!\cdot 7!^7=7!^8.$$ To see where that inequality comes from, multiply out the factorials so that $$8^7=8\cdot 8\cdot 8\cdot 8\cdot 8\cdot 8\cdot 8>7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=7!.$$

Answer 4

inequality equivalent to $$(7!)<(8!)^{7/8}$$ $$(8!)^{1/8}(7!)<(8!)^{1/8}(8!)^{7/8}$$ $$(8!)^{1/8}(7!)<(8!)$$ $$(8!)^{1/8}(7!)<8(7!)$$ $$(8!)^{1/8}<8$$

Answer 5

We could generalize the matter this way. We can solve

$$ ( \ n! \ )^{1/n} \ = \ x^{\frac{1}{n+1}} \ \ \Rightarrow \ \ x \ = \ ( \ n! \ )^{\frac{n+1}{n}} \ = \ ( \ n! \ ) \ · \ ( \ n! \ )^{\frac{1}{n}} $$

and would wish to claim that

$$ ( \ n! \ ) \ · \ ( \ n! \ )^{\frac{1}{n}} \ < \ \ ( n + 1 )! \ \ \Rightarrow \ \ ( \ n! \ )^{\frac{1}{n}} \ < \ n + 1 \ \ , \ \ \text{for} \ n \ \ge \ 1 \ \ . $$

Taking the logarithm of both sides, we would have

$$ \frac{1}{n} \ \log( \ n! \ ) \ < \ \log(n + 1) $$ $$ \Rightarrow \ \log \ n \ + \ \log (n-1) \ + \ \ldots \ + \ \log \ 2 \ + \ \log \ 1 \ < \ n \ \log (n+1) \ \ . $$

This last inequality is certainly true for $ \ n \ \ge \ 1 \ $ , so it follows that $$ \ ( \ n! \ )^{\frac{1}{n}} \ · \ ( \ n! \ ) \ < \ n + 1 \ · \ ( \ n! \ ) \ \ \Rightarrow \ \ ( \ n! \ )^{\frac{n+1}{n}} \ < \ (n + 1)! $$ $$ \Rightarrow \ \ ( \ n! \ )^{\frac{ 1}{n}} \ < \ [ \ (n + 1)! \ ]^{\frac{1}{n+1}} \ \ , \ \text{for} \ n \ \ge \ 1 \ \ . $$

[This is related to a number of the approaches given by other responders.]

A more sophisticated (even high-falutin') way of proving this would be to say that the proposition follows from the fact that $ \ \frac{\log \ \Gamma(x + 1) }{x} \ $ is known to be an increasing function for $ \ x \ \ge \ 1 \ $ .