A tangled nest we weave

We can rewrite $S$ as

$\pi^{\frac{1}{2}} * e^{\frac{1}{6}} * \pi^{\frac{1}{24}} * e^{\frac{1}{120}} * \pi^{\frac{1}{720}} * ...... = \pi^{M} * e^{N}$

where $M = \displaystyle\sum_{k=1}^{\infty} \dfrac{1}{(2k)!} = \cosh(1) - 1$ and

$N = \displaystyle\sum_{k=1}^{\infty} \dfrac{1}{(2k + 1)!} = \sinh(1) - 1$ .

(See here for a discussion of the Taylor series for hyperbolic functions.)

Thus $S = \pi^{\cosh(1) - 1} * e^{\sinh(1) - 1} = 2.2186121.....$ , and so $\lfloor 1000*S \rfloor = \boxed{2218}$ .

My solution may not be impressive for mathematicians but I used the given information to the best.

repeated exponentiation give u the general term as

S^(n!) = pi^(Sum(1/2n!)) * e^(Sum(1/(2n+1)!) //n from 1 to infinity

I couldn't find the hyperbolic function-equivalent for days then I looked at the answer we need here, 1000*S

and after n=7, 1/2n! etc won't affect the Sum much which means that for 1000*S value you can just calculate Powers of pi and e upto n=8. i.e., 1/2! + 1/4! +.... +1/16! , similarly for e, 1/3! + 1/5! + 1/7! ...+1/17!

and calculate the approximate answer.

And Bingo!! It worked :)