If $n,m \in \mathbb{N} - \{0\}$ such that $n,m \geq 2$ and $n > m$ then $n^{\frac{1}{n!}} < m^{\frac{1}{m!}}$. As the title says, I'm trying to show that if $n,m \in \mathbb{N} - \{0\}$ such that $n,m \geq 2$ and $n > m$ then $n^{\frac{1}{n!}} < m^{\frac{1}{m!}}$.
I've tried using logarithms, but it is notoriously difficult since logarithm is an increasing function, therefore it doesn't flip the inequality between $n$ and $m$ the way I need it to. Here's what I mean
$$n > m \Rightarrow n! > m! \Rightarrow \frac{1}{n!} < \frac{1}{m!}$$
Since $n > m \Rightarrow \ln (n) > \ln (m)$
$$\Rightarrow \frac{1}{m!} \ln (n) > \frac{1}{n!} \ln (n) > \frac{1}{n!} \ln (m)$$
Here's where I get stuck. I naturally thought of using logarithms since they let us treat exponentiation as multiplication, but I can't seem to get anywhere. Is there a better way to approach this problem?
Thank you for your time.
 A: Maybe I’m missing something, but if you take the function
$$X^{1/X!}$$
And take log then you get
$$\log(X)/X!$$
Since $\log(X)/X$ is a decreasing function for $X$ above $e$ (you can check by differentiating), this function decreases as $X$ increases past $e$.  It also decreases as $X$ goes from $2$ to $3$ as you can see by hand.
So $\log(n)/n!$ is less than $\log(m)/m!$
If you then take exponentials and rely on $\exp$ being an increasing function then you are home.
A: We will prove this for two consecutive integers first and then perform induction over the larger of the two integers. Say we are working with $a+1$ and $a.$ We want to prove that $$(a+1)^{\frac{1}{(a+1)!}}<a^{\frac{1}{a!}}.$$ This is equivalent to $$1<(a^a - 1)a,$$ which is true because $$2\le (a^a -1)a$$ due to the fact that $a\ge 2.$ Now we wish to show that if $n>m$ for fixed $m$ and $n^{\frac{1}{n!}}<m^{\frac{1}{m!}},$ then $$(n+1)^{\frac{1}{(n+1)!}}<m^{\frac{1}{m!}}.$$ We get what we want because
\begin{align*}
n^{\frac{1}{n!}}&<m^{\frac{1}{m!}}\\
n&<m^{\frac{n!}{m!}}\\
n^{n+1}&<m^{\frac{(n+1)!}{m!}}.
\end{align*}
It suffices to prove that $$n+1<n^{n+1},$$ which we already showed except with $a$ instead of $n$ earlier.
A: Let's define $a(n)=n^{\frac{1}{n!}}$. We want to show that this sequence is decreasing since $n>2$. To obtain this result let's consider the fraction $\frac{a(n+1)}{a(n)}$. We have
$$\frac{(n+1)^{\frac{1}{(n+1)!}}}{n^{\frac{1}{n!}}}=\left(\frac{(n+1)^{\frac{1}{n+1}}}{n}\right)^{\frac{1}{n!}}$$
For $n>2$: $\sqrt[n+1]{n+1}<n$ and $\frac{1}{n!}>0$, so $\frac{a(n+1)}{a(n)}<1$, which implies that the sequence is decreasing.
