Limit with a big exponentiation tower 
Find the value of the following limit:
  $$\huge\lim_{x\to\infty}e^{e^{e^{\biggl(x\,+\,e^{-\left(a+x+e^{\Large x}+e^{\Large e^x}\right)}\biggr)}}}-e^{e^{e^{x}}}$$
(original image)

I don't even know how to start with. (this problem was shared in Brilliant.org)
Some of the ideas I tried is to take the natural log of this expression and reduce it to $\ln(a/b)$ then use L'Hopital's but that made it false!!
I know the value of the limit it is $e^{-a}$ but please how to prove it?

Let $ A=\exp\left(x+e^{-\left(a+x+e^x+e^{e^x}\right)}\right) $ and $B=\exp(x) $. Then we can easily conclude that $A/B$ tends to $1$, but a little more analysis allows us to infer that $A-B$ also tends to $0$.
We have $$A-B=B\cdot\frac{\exp(\log A-\log B) - 1}{\log A-\log B} \cdot(\log A-\log B) $$ and the middle fraction tends to $1$ and hence the limit of $A-B$ is same as that of $$B(\log A-\log B) $$ ie $$e^x\cdot e^{-\left(a+x+e^x+e^{e^x}\right)}$$ and thus $A-B$ tends to $0$.
It follows that $C/D\to 1$ where $C=e^A, D=e^B$. Applying same idea we note that limit of $C-D$ is same as that of $$D(\log C-\log D)=D(A-B) $$ which is same as that of $$DB(\log A-\log B) $$ ie $$e^{e^x} e^x\cdot e^{-\left(a+x+e^x+e^{e^x}\right)}$$ Thus we conclude that $C-D\to 0$.
Next let $E=e^C, F=e^D$ then $E/F\to 1$ and limit of $E-F$ is same as that of $$F(\log E-\log F) =F(C-D) $$ which is same as that of $$FDB(\log A-\log B) $$ ie $$e^{e^{e^x}} e^{e^x} e^x\cdot e^{-\left(a+x+e^x+e^{e^x}\right)}$$ The desired limit is then $e^{-a} $.
