Conjecture: $\lim\limits_{x\to 0}(x!\,x!!\,x!!!\,x!!!!\cdots )^{-1/x}\stackrel?=e$ Well, it's a conjecture so let me propose it:
$$\lim_{x\to 0}(x!\,x!!\,x!!!\,x!!!!\cdots)^{-1/x}\stackrel?=e$$
Where I use desmos notation and $x!! := ((x!)!,x!!!=(((x!)!)!)$

It seems so hard that I haven't any clue to show it. I already know that
$$x!^{\frac{-1}{x}}=e^{\gamma}$$
Perhaps we can use the Weierstrass factorization theorem and compute it. So, how to (dis)prove it?
Further investigation :
If we supposed that the following functions are convex on $(0,1)$:
$$a_1(x)=x!,a_2=x!!,\cdots,a_n=x!!\cdots !$$
Then we can rewrite the conjectured limit as :
$$L=\lim_{x\to 0}\left((a_1'(0)x+1)(a_2'(0)x+1)\cdots(a'_n(0)x+1)\cdots\right)^{\frac{-1}{x}}=^?e$$
Where :
$$a_1'(0)=\gamma,a_2'(0)=(\gamma-1)\gamma,a_3'(0)=-(\gamma-1)^2\gamma,a_4'(0)=(\gamma-1)^3\gamma,\cdots\tag{I}$$
Update :
It seems we have :
$$\lim_{x\to 0^+}\left(x!x!!x!!!x!!!!x!!!!!...\right)^{\frac{2^{x}-1}{x^{2}}}=1/2$$
Update $2$ :
Using $I$ and the fact that (see Robjohn's answer) :
$$\sum_{n=1}^\infty\gamma(1-\gamma)^{n-1}=1$$
And :
Let $x_i\in[1-1/n,1]$ where $n\geq M$ two natural numbers large enought then we have :
$$\sum_{i=1}^{n}x_i-(n-1)\leq \prod_{i=1}^{n}x_i\leq \sum_{i=1}^{n}x_i-(n-1)+\frac{1}{2n}$$
We have after simplification :
$$\left(-x+1\right)^{\left(-\frac{1}{x}\right)}<L<\left(-x+1+\frac{1}{2n}\right)^{\left(-\frac{1}{x}\right)}$$
now let $n\to \infty$ and $x\to 0$ we get the result .
Ps:It's a try and I think it should be clearing a little (the LHS seems dubious) and the credit come back to @Robjohn.
 A: If we consider the function $$f(x) = \sum_{k=1}^\infty \log(x(!^k))$$
then the question is equivalent to prove that $f'(0)=-1$.
We have (assuming that we can differentiate term by term) that$$f'(x) = \sum_{k=1}^\infty\frac{d}{dx} \log(x(!^k))= \sum_{k=1}^\infty\frac{\frac{d}{dx} x(!^k)}{x(!^k)}$$
Since $0(!^k)=1$ we have
$$f'(0)=\sum_{k=1}^\infty \left.\frac{d}{dx}x(!^k)\right\vert_{x=0}$$
Using $\frac{d}{dx}x! = x!\psi(1+x)$ and a repeated application of the chain rule yields $$\frac{d}{dx}(x(!)^k) = x! x(!^2)x(!^3)\cdots x(!^k) \psi(1+x)\psi(1+x(!^1))\psi(1+x(!^2))\cdots\psi(1+x(!^{k-1}))$$
where $\psi$ is the digamma function.
Using again that $0(!^k) = 1$, that $\psi(1) = -\gamma$ and $\psi(2) = 1-\gamma$ we see that $\left.\frac{d}{dx}x(!^k)\right\vert_{x=0} = -\gamma\,(1-\gamma)^{k-1}$ and then we have $$f'(0)=\sum_{k=1}^\infty-\gamma\, (1-\gamma)^{k-1} =-\gamma\,\frac{1}{1-(1-\gamma)} = -1$$
A: At first Mathematica appears to support your conjecture numerically, e.g. using $x!=\Gamma(x+1)$ then
$$\underset{x\to 0}{\text{lim}}(\Gamma (x+1) \Gamma (\Gamma (x+1)+1) \Gamma (\Gamma (\Gamma (x+1)+1)+1) \Gamma (\Gamma (\Gamma (\Gamma (x+1)+1)+1)+1) \Gamma (\Gamma (\Gamma (\Gamma (\Gamma (x+1)+1)+1)+1)+1) \Gamma (\Gamma (\Gamma (\Gamma (\Gamma (\Gamma (x+1)+1)+1)+1)+1)+1) \Gamma (\Gamma (\Gamma (\Gamma (\Gamma (\Gamma (\Gamma (x+1)+1)+1)+1)+1)+1)+1))^{-1/x}$$
You can try it yourself if you have Mathematica (thanks to @KStarGamer for the improved code):
Limit[Product[Nest[Gamma[# + 1] &, x, n], {n, 1, 7}]^(-1/x), x -> 0]

In the above example we obtain
$$\exp \left(\gamma  \left(7-21 \gamma +35 \gamma ^2-35 \gamma ^3+21 \gamma ^4-7 \gamma ^5+\gamma ^6\right)\right),$$
and the exponent does seem to tend to $1$.
The coefficients in the exponent polynomial in $\gamma$ appear to be those of A007318 (binomial coefficients !) So the general case could be binomial in some way.
It looks like the general case is
$$e^{1-(1-\gamma )^n}\to e^1$$
as $n\to\infty$.
For a proof you may be able to use the fact that $$\Gamma(x+1)=1-\gamma x + O(x^2).$$
