Not clear in one step in the inequality for the proof LCM$(1,2,\dots,n) < 3^n$ I believe that I am following Hanson's proof up until this point (page 36):

$$C(n) < \dfrac{(en)^{k-1+1/(a_{k+1}+1)}w^n}{(a_1)^{(a_1-1)/a_1}(a_2)^{(a_2-1)/a_2}\dots(a_m)^{(a_k-1)/a_k}}$$

where $e$ is Euler's constant, $a_1=2$, $a_{i+1} = a_1 a_2 \dots a_i + 1$ so that:


*

*$a_2 = 3$

*$a_3 = 7$
$k$ is the value where $a_k \le n < a_{k+1}$
$w^n$ is not important for my question but it is defined as $w = a_1^{1/a_1}a_2^{1/a_2}\dots$
I am unclear at this step where:

$$C(n) < \dfrac{(en)^{k-1+1/(a_{k+1}+1)}w^n}{(a_1)^{(a_1-1)/a_1}(a_2)^{(a_2-1)/a_2}\dots(a_m)^{(a_k-1)/a_k}}< e^{k - 3/2}n^{k - 3/2}w^n$$

where $k > 2$ since $n < a_1 a_2 \dots a_k$
Here's what I understand:


*

*Since each $a_i > 1$, I am clear why:



$$\dfrac{(en)^{k-1+1/(a_{k+1}+1)}w^n}{(a_1)^{(a_1-1)/a_1}(a_2)^{(a_2-1)/a_2}\dots(a_m)^{(a_k-1)/a_k}}< e^{k - 1 + 1/(a_{k+1}+1)}n^{k - 1 + 1/(a_{k+1}+1)}w^n$$



*

*If he meant to use $k - \frac{3}{4}$, then I would be clear since:



$$(en)^{k-1+1/(a_{k+1}+1)}w^n < e^{k - 3/4}n^{k - 3/4}w^n$$



*

*If $k - \frac{3}{2}$ is correct, then I would appreciate help in understanding how this follows.


It is not obvious to me that this is true or why it is true. 
 A: First of all, from your previous question, I think (2.6) should be
$$C(n) < \dfrac{(en)^{k-1+1/(a_{k+1}\color{red}{-}1)}w^n}{a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k}}\lt  e^{k - 3/2}n^{k - 3/2}w^n$$
(see the minus sign in red in the numerator)
To prove 
$$\dfrac{(en)^{k-1+1/(a_{k+1}-1)}w^n}{a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k}}\lt  e^{k - 3/2}n^{k - 3/2}w^n\tag1$$
it is sufficient to prove
$$(en)^{k-1+1/(a_{k+1}-1)-(k-3/2)}\lt  a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k},$$
i.e.
$$(en)^{1/2+1/(a_{k+1}-1)}\lt  a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k}$$
Since $n\le a_1a_2\cdots a_k$, it is sufficient to prove
$$(ea_1a_2\cdots a_k)^{1/2+1/(a_{k+1}-1)}\lt  a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k},$$
i.e.
$$e^{1/2+1/(a_{k+1}-1)}\lt a_1^{1/2-1/a_1-1/(a_{k+1}-1)}a_2^{1/2-1/a_2-1/(a_{k+1}-1)}\cdots a_k^{1/2-1/a_k-1/(a_{k+1}-1)},$$
i.e.
$$e^{1/2+1/(a_{k+1}-1)}2^{1/(a_{k+1}-1)}\lt a_2^{1/2-1/a_2-1/(a_{k+1}-1)}\cdots a_k^{1/2-1/a_k-1/(a_{k+1}-1)}$$
Since $a_i^{1/2-1/a_i-1/(a_{k+1}-1)}\gt 1$ for $2\le i\le k-1$,
It is sufficient to prove 
$$e^{1/2+1/(a_{k+1}-1)}2^{1/(a_{k+1}-1)}\lt a_k^{1/2-1/a_k-1/(a_{k+1}-1)}\tag2$$
for $k\ge 3$.
Note here that seeing $(2)$ as an inequality on $k$, we see that the LHS of $(2)$ is decreasing and that the RHS of $(2)$ is increasing, so it is sufficient to prove $(2)$ for $k=3$, which is equivalent to
$$2e^{22}\lt 7^{14}$$
which is indeed true.
Therefore, $(1)$ is true.
