Does this $a_n=(1-\frac{1}{2!})^{(\frac{1}{2!}-\frac{1}{3!})^{\ldots^{(\frac{1}{n!}-\frac{1}{(n+1)!})}}}$ have a finit limit? My question here is related to telescopic sum using factorial and it is related to my question here, I have computed some values of $a_n=(1-\frac{1}{2!})^{(\frac{1}{2!}-\frac{1}{3!})^{\ldots^{(\frac{1}{n!}-\frac{1}{(n+1)!})}}}$ for odd parity and even parity but it is not fixed for example for $n=2$ we have $0.793700$ and it decreases for $n=4$ to $0.77982$, now for $n=3$ we have $0.5465$ and it increases for $n=5$ to $0.54876$ , it seems increasing for odd parity and decreasing for even parity iteration. Now I have looked to all given answers here but I can't juge whether that sequence converges or not by means it has a limit or not?


  
    My question here is:
    Is this $a_n=(1-\frac{1}{2!})^{(\frac{1}{2!}-\frac{1}{3!})^{\ldots^{(\frac{1}{n!}-\frac{1}{(n+1)!})}}}$ have a finit limit ?
  


Note The motivation of this question is looking to the behavior of the Gamma function in the power telescoping sum.
We have similar behavior to the sequence that you linked to, except the limiting values are different:  For even $n$, it is $$a_n \to 0.77954333600168773503298455024204190801488463615921\ldots,$$ and for odd $n$, it is $$a_n \to 0.54877354704085687513069922740691455562600046738030\ldots.$$  The number of correct decimal places increases slightly faster than quadratically in $n$; i.e., if $\epsilon(n)$ is the absolute error as a function of $n$, then $$-\log \epsilon(n) \sim O(n^{k}),$$ where I estimate $k$ to be approximately $2.3$, certainly greater than $2$ but less than $2.4$.  I do not recognize these constants as having a closed form.
I don't think this really counts as a proof, but I made some code in Matlab code to investigate this.

n=input('please input a value for n ');

for m=1:n;

for I=1:m;

    b(m+1)=1;

    b(m-I+1)=(1/factorial(m-I+1)-1/factorial(m-I+2))^b(m-I+2);

    a(m)=b(1);

end


end

for I=1:n;

fprintf('a(%.f)=%.8f \n',I,a(I));

end

Inputting n=12 outputs
a(1)=0.50000000 
a(2)=0.79370053 
a(3)=0.54650798 
a(4)=0.77982290 
a(5)=0.54876028 
a(6)=0.77954370 
a(7)=0.54877354 
a(8)=0.77954334 
a(9)=0.54877355 
a(10)=0.77954334 
a(11)=0.54877355 
a(12)=0.77954334 
So while it seems that $a_{2n}$ and $a_{2n-1}$ each converge, they don't converge to the sam value.

The thing is that as n tends to infinity $(\frac{1}{(n)!}-\frac{1}{(n+1)!})$ rapidly tends to 0, so $(\frac{1}{(m-n)!}-\frac{1}{(m+1-n)!})^{(\frac{1}{(n+1)!}-\frac{1}{(n+2)!})}$ quickly tends to 1.