Showing the Fibonacci inequality $f_1^{f_1}f_2^{f_2}f_3^{f_3}\cdots f_n^{f_n}\leq f_1!f_2!f_3!\cdots f_n!\;e^{({f_{n+2}-n-1)}}$ without induction. Today I saw the following 2 beautiful inequalities on a facebook page.

$$1^12^23^34^4\cdots n^n\leq 1!2!3!4!\cdots n!e^{\frac{n(n-1)}{2}}$$
$$f_{1}^{f_{1}}f_{2}^{f_{2}}f_{3}^{f_{3}}\cdots f_{n}^{f_{n}}\leq f_{1}!f_{2}!f_{3}!\cdots f_{n}!e^{({f_{n+2}-n-1)}}$$

Here $f_{n}$ denotes the fibonacci numbers $f_{1}=f_{2}=1$ and $f_{n+2}=f_{n+1}+f_{n}$ for $n\in N$
Here is How I proved the first one.
Method 1:-
$$1^12^23^34^4...n^n\leq 1!2!3!4!...n!e^{\frac{n(n-1)}{2}}$$
$$\implies\bigg(\frac{1^1}{1!}\bigg)\bigg(\frac{2^2}{2!}\bigg)\bigg(\frac{3^3}{3!}\bigg)...\bigg(\frac{n^n}{n!}\bigg)\leq e^{\frac{n(n-1)}{2}}$$
$\implies \displaystyle\prod_{x=1}^{n} \frac{x^x}{x!}\leq 
e^{\frac{n(n-1)}{2}}$
Now taking logarithm on both sides of inequality.
$\implies \displaystyle\sum_{x=1}^{n} \ln\bigg(\frac{x^x}{x!}\bigg)\leq 
{\frac{n(n-1)}{2}}$
$\implies \displaystyle\sum_{x=1}^{n} \ln\bigg(\frac{x^x}{x!}\bigg)\leq 
{\frac{n(n+1)}{2}}-n$
$\implies \displaystyle\sum_{x=1}^{n} \ln\bigg(\frac{x^x}{x!}\bigg)\leq 
\bigg(\sum_{x=1}^{n} x \bigg)-n$
$\implies \displaystyle \ \sum_{x=1}^{n} x -\displaystyle\sum_{x=1}^{n} \ln\bigg(\frac{x^x}{x!}\bigg)\geq n$
$\implies \displaystyle \ \sum_{x=1}^{n} x + \ln\bigg(\frac{x!}{x^x}\bigg)\geq n$
$$\implies 1+1.30+1.49+1.63+1.74+1.82+1.90+1.97+2.02+...\geq n$$
$\implies 1+(1+0.30)+(1+0.49)+(1+0.63)+(1+0.74)+...\geq  {\smash[b]{1+\! \underbrace{1+1+\cdots1\,}_\text{$n$ times}}}$
This proves our first inequality.
Method 2:-
Using $\frac{n^n}{n!}\leq e^{n-1}$ for $n\in N$.
$\frac{1^1}{1!}\leq e^{0}$
$\frac{2^2}{2!}\leq e^{1}$
$\frac{3^3}{3!}\leq e^{2}$
$\vdots\\$
$\frac{n^n}{n!}\leq e^{n-1}$
Now multiply all above inequalities we get,
$\implies \displaystyle\prod_{x=1}^{n} \frac{x^x}{x!}\leq 
e^{\frac{n(n-1)}{2}}$
Using the same approach for $2^{nd}$ inequality we get,
$$\sum_{x=1}^{n} \ln\bigg(\frac{{f_{x}}^{f_{x}}}{f_{x}!}\bigg)\leq f_{n+2}-n-1$$
Now I got stuck on this step.

How can we prove the $2^{nd}$ inequality. Moreover can we show without numerically calculating(as I do above) that $\displaystyle \ \sum_{x=1}^{n} x + \ln\bigg(\frac{x!}{x^x}\bigg)\geq n$

 A: The case $n=0$ reduces to $1\le e^{f_2-1}$, as its left-hand side is an empty product. Increasing $n$ from $k$ to $k+1$ multiplies the left- (right)-hand side by $f_{k+1}^{f_{k+1}}$ ($f_{k+1}!e^{f_{k+1}-1}$), so just prove $f_{k+1}^{f_{k+1}}\le f_{k+1}!e^{f_{k+1}-1}$. Indeed, you can prove by induction $m^m\le m!e^{m-1}$ for $m\ge1$. In particular,$$u_m:=\frac{m!e^{m-1}}{m^m}\implies u_1=1,\,\frac{u_{m+1}}{u_m}=\frac{e}{(1+1/m)^m}>1.$$
A: Are known sums of the Fibonacci numbers
$$F_{n+2}-1 =\sum\limits_{k=1}^n F_n\tag1$$
and the Stirling's inequality
$$m!\ge\sqrt{2\pi m}\, m^m\,e^{-m},\quad m\in \mathbb N.\tag2$$
Then, taking in account $(2),$
\begin{align}
&1!\,e^{1-1} = 1 = 1^1,\\[4pt]
&2!\,e^{2-1} \ge \sqrt{4\pi}\, 2^2 e^{-2+2-1} \ge 2^2,\dots,\\[4pt]
&m!\,e^{m-1} \ge \sqrt{2\pi m}\, m^m e^{-m+m-1} \ge m^m,\dots,\\[4pt]
\end{align}
$$m!\,e^{m-1} \ge m^m,\quad m\in\mathbb N.\tag3$$
Finally,
$$e^{F_{n+2}-1-n}\prod\limits_{k=1}^n F_k! = \prod\limits_{k=1}^n F_k!e^{F_k-1}
\ge \prod\limits_{k=1}^n F_k^{F_k}.$$
The first inequality allows the same approach.
