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.

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.

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,$$u_m:=\frac{m!e^{m-1}}{m^m}\implies u_1=1,\,\frac{u_{m+1}}{u_m}=\frac{e}{(1+1/m)^m}>1.$$