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.
 A: 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.
A: 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.
