How to show that the inequality with exponents is true In part of a question in my homework I'm doing, I need to show that:
$$ \log(n) > \sqrt [n]{n} $$
Is this a correct a correct manipulation:
$$n > e^\sqrt [n]{n}$$
Thank you in advance!
edit:
that's a good point that it works only for n>4, and I forgot to mention that n is a natural number ({1,2,...})
 A: (In the true spirit of stackexchange:)
Yes.
A: Well, $\log(1) > \sqrt[1]1$ is wrong, and as well n=2,3,4.  So let's suppose $n \geq 5$ (it holds for $n=5$).
Put it as 
$$
(\log (n))^n
> n
$$
Then use induction,
$$
%(\log (n+1))^{n+1} >=(\log (n) + \log(1 + 1/n))^{n+1}
(\log (n+1))^{n+1} >(\log (n))^{n+1} = \log(n) (\log (n))^{n} \geq  n \log(n)
$$
Where the last one holds by induction.
Now show 
$$\log(n) > 1 + 1/n$$ which holds for $n=5$, and then also for larger n since the log is increasing with n and $1/n$ is decreasing with n. 
So one can continue 
$$
%(\log (n+1))^{n+1} >=(\log (n) + \log(1 + 1/n))^{n+1}
(\log (n+1))^{n+1}  \geq  n \log(n)
> n (1 + 1/n) = n+1
$$
Done. $\quad \quad \Box$
A: Here's a proof by induction since $n \in \mathbb{Z^{+}}$
$n=1$, we have:
LHS = $\log 1$ = 0 
RHS = $1 > $LHS hence the statement is incorrect.
A: True, lets manipulate the inequality $ \ln(n) > n^{\frac{1}{n}}$  which can be turn into $\ln(\ln(n)) > \frac{1}{n} \ln(n)$ (because $\ln(a^b) = b \ln(a)$)
And then multiply by $n$ we get that $n \ln(\ln(n)) > \ln(n)$ when $\ln(\ln(n)) > 1$ we can reduce it to $n > \ln(n)$ which can be turn into $e^n > n$ which is true and can be proven by induction $n=1$ => $e^1 \approx 2.718 >1 $ now assume its true for $e^{k} > k$ prove that $e^{k+1} > k+1$ which is $e^{k}*e >k+1$ which is $k*e > k+1$ using the assumption,which can be reduces to $2k>k+1$ and its true when $k>1$, last step is to see when $\ln(\ln(n)) >1$   which is just when $n > e^{e^{1}} \approx 15.15$ ,so the inequality is true when $n >15$ and when $n \leq 15$ a check is enough to show that its true when $n >4$
A: Andreas said to holds for $5$ , which is not true ; Ahmad has the correct concepts but uses the $\log_e$ while the question uses base $10....n ≥16$ for the inequality to hold , then one can take the derivative of both sides and see the left increases and right decreases for $n ≥ 16 \cdots$ thus it holds for $n ≥ 16$
