Is $\left(1-\dfrac1n\right)^{a_n}$ increasing? Let $a_n=\left\lfloor\sqrt{2n-2}\right\rfloor+1$. Prove that this holds for $n\in\Bbb N$.

$$\left(1-\frac{1}{n}\right)^{a_n}\le\left(1-\frac{1}{n+1}\right)^{a_{n+1}}$$

My try:

If $n=1$ then $a_n=a_1=\left(1-\frac11\right)^{a_1}=0$ and
$a_{n+1}=a_2=\left(1-\frac12\right)^{a_2}$, which is $1/2$ or $1/4$,
depending on $a_2$. So the statement is true for $n=1$.
Let $n$ the least natural number that does not hold the inequality. [This is a subtle way to use induction on $n$].
Using the sum of geometric progressions, we obtain: $$\sum_{k=0}^{a_n} \left(1-\frac{1}{n}\right)^k=\frac{1-\left(1-\dfrac{1}{n}\right)^{a_n+1}}{1-\left(1-\dfrac{1}{n}\right)}=n\left[1-\left(1-\dfrac{1}{n}\right)^{a_n+1} \right]$$
therefore,
$$1-\left(1-\dfrac{1}{n}\right)^{a_n+1}=\frac1n \sum_{k=0}^{a_n} \left(1-\frac{1}{n}\right)^k$$
And similarly,
$$1-\left(1-\dfrac{1}{n+1}\right)^{a_{n+1}+1}=\frac1{n+1} \sum_{k=0}^{a_{n+1}} \left(1-\frac{1}{n+1}\right)^k$$
Since $n$ does
not hold the statement,
$$\frac1{n+1} \sum_{k=0}^{a_{n+1}} \left(1-\frac{1}{n+1}\right)^k>\frac1n \sum_{k=0}^{a_n}  \left(1-\frac{1}{n}\right)^k$$

But I dont know how to go on. I'd like a proof like this. No limits, continuity or mention to number $e$.
Perhaps the statement hols only for large enough $n$. If it holds for $n\ge 100$, this would work for me, too.
 A: Taking the logarithm to both sides of
$$\left(1-\frac{1}{n}\right)^{a_n}\le\left(1-\frac{1}{n+1}\right)^{a_{n+1}} \tag{1}$$
you get
$$\frac{\lfloor\sqrt{2n-2}\rfloor+1}{\lfloor\sqrt{2n}\rfloor+1}\overset{?}{{>}} \frac{\ln\Big(1-\frac{1}{n+1}\Big)}{\ln\Big(1-\frac{1}{n} \Big)}\tag{2}$$
Note that since $(1-\frac{1}{k})$ is less than one,  $\ln(1-\frac{1}{k})$ is negative therefore the direction of the inequality has been changed.
We will have that $\lfloor\sqrt{2n-2}\rfloor$ is not equal $\lfloor\sqrt{2n}\rfloor$ only when $2n$ is a perfect square:
$$\begin{matrix}
\lfloor\sqrt{2n-2}\rfloor=k-1 & \text{when } 2n =k^2 \\
\lfloor\sqrt{2n-2}\rfloor=\lfloor\sqrt{2n}\rfloor & \text{otherwise}
\end{matrix}$$
therefore we will have
$$1\overset{?}{{>}} \frac{\ln\Big(1-\frac{1}{x+1}\Big)}{\ln\Big(1-\frac{1}{x} \Big)}\tag{a}$$
$$\frac{k}{k+1}\overset{?}{{>}} \frac{\ln\Big(1-\frac{2}{k^2+2}\Big)}{\ln\Big(1-\frac{2}{k^2} \Big)}\tag{b}$$
The relation $(a)$ holds for every real $n$ and therefore also for every integer $n$ since the argument of the logarithm to the denominator is greater than the argument of the logarithm to the numerator. While $(b)$ didn't holds for any real $k$. So whenever $2n$ is a perfect square $(1)$ it is false
$$\left(1-\frac{2}{k^2}\right)^{k}>\left(1-\frac{2}{k^2+2}\right)^{k+1} \Rightarrow \left(1-\frac{2}{2n}\right)^{\lfloor\sqrt{2n-2}\rfloor+1}>\left(1-\frac{2}{2n+2}\right)^{\lfloor\sqrt{2n}\rfloor+1}$$
