Why is $5^n*(1-1/n)^n > 5^m *(1-1/n)^m $ for $m1?$ Let $n$ be a positive integer greater than $1.$ If $m<n,$ is a positive integer less than $n,$ it seems as if the following inequality is true:  
$$5^n*(1-1/n)^n > 5^m*(1-1/n)^m.$$   
How would I go about showing this is true, if it is? Can we get a more approximation of how much these two expressions differ? The obvious thing would be of course to divide both sides and prove that $$5^{n-m} (1-1/n)^{n-m} < 1.$$ But this seems just as hard as the previous inequality. One idea would be to maybe prove the inequality 
$$5^n(1-1/n)^n > 5^m(1-1/n)$$ and then we must show that 
$$5^{n-m} (1-1/n)^{n-1} > 1.$$  
Any solution would be welcome.
 A: You have to prove, according to what you wrote, that
$$n>m\implies5^n\left(1-\frac1n\right)^n>5^m\left(1-\frac1n\right)^m\iff5^{n-m}\left(1-\frac1n\right)^{n-m}>1\iff$$
$$\left(5-\frac5n\right)^{n-m}>1\iff 5-\frac5n>1$$
and since the last inequality is trivialy true for $\;n\ge2\;$, we're done.
A: The idea, as Dave writes is that when $x > 1$ then $x^n > x^m$ when $n > m$. In this case
$$ x = 5(1 - 1/n) = 5 - 5/n. $$
This is $>1$ when
\begin{align}
5 - 5/n &> 1 \\
5n - 5 &> n \\
4n - 5 &> 0 \\
4n &> 5 \\
n &> 5/4.
\end{align}
Well, in this case we know this is true because $n \ge 2$.

The difference, since you've asked is
$$ (5 - 5/n)^n - (5 - 5/n)^m = (5 - 5/n)^m ((5 - 5/n)^{n - m} - 1). $$
Now it depends on what $m$ and $n$ are. Let's say $m$ is roughly linear in $n$. Say $m = \alpha n - k$ where $0 \le \alpha  \le 1$.
If $\alpha < 1$ then $(5 - 5/n)^{n - m} \gg 1$ and the difference is approximately
$$ (5 - 5/n)^n \approx 5^n e^{-1}. $$
If $\alpha = 1$, the difference is approximately
$$ 5^n (1 - 5^{-k}) e^{-1}. $$
For instance when $\alpha = 1$ and $k = 1$ the difference is approximately
$$ \frac45 5^n e^{-1}. $$
