Convergence of $a_n=(1-\frac12)^{(\frac12-\frac13)^{...^{(\frac{1}{n}-\frac{1}{n+1})}}}$ let us moving to telescopic sum using exponent ,Assume we have this sequence: $a_n=(1-\frac12)^{(\frac12-\frac13)^{...^{(\frac{1}{n}-\frac{1}{n+1})}}}$ with $n\geq1$ , this sequence can be written as power of sequences  : ${x_n} ^ {{{y_n}^{c_n}}^\cdots} $ such that all them value are in $(0,1)$, I want to know if the titled sequence should converge to $1$ ? and how we can evaluate it for $n$ go to $\infty$  ? 
 A: Numeric calculation of the sequence $\{a_n\}_{n \ge 1}$ suggests that the terms are bounded, but alternate between approximately $$0.56778606544394002098000796382530333102219963214866$$ and $$0.85885772008416606762434379473241623070938618180813,$$ but I do not have a proof.  This convergence is extremely rapid, and the alternating nature suggests that it is important to look at even and odd $n$ separately.
A: This only shows that the limit cannot be $1$.
Note that $a_n=(1/2)^{(1/6)^{(1/12)^\cdots}}$, where the "$\cdots$" are meant to terminate at the exponent $1/(n(n+1))$.
As a general rule, if $0\lt r\lt1$ and $0\lt a\lt b\lt1$, then $0\lt r\lt r^b\lt r^a\lt1$.  It follows that 
$$0\lt(1/12)\lt(1/12)^{(1/20)^\cdots}\lt1$$ 
and thus also that 
$$0\lt(1/6)\lt(1/6)^{(1/12)^{(1/20)^\cdots}}\lt(1/6)^{(1/12)}\lt1$$ 
so that, finally,
$$0.5504566141\approx(1/2)^{(1/6)^{(1/12)}}\lt(1/2)^{(1/6)^{(1/12)^\cdots}}\lt(1/2)^{(1/6)}\approx0.89089871814$$
These bounds accord with what heropup found.
A: It is perhaps interesting to note that the behaviour noted by previous posts is true for a wide class of functions defined by towers.
For all positive integers $i$ let $u_i$ be any real numbers such that $1>u_i>0$. Define $$a(n)=u_1^{u_{2}^{...^{u_n}}},b(n)=u_2^{u_{3}^{...^{u_n}}}.$$
Lemma 1 $$a(1)<a(3)<a(5)<a(7)...$$ $$1>a(2)>a(4)>a(6)>a(8)...$$
Proof $$a(N+2)-a(N)=u_1^{b(N+2)}-u_1^{b(N)}.$$
Therefore $a(N+2)-a(N)$ and $b{(N+2)}-b(N)$ have opposite signs. 
Now $b(N+2)-b(N)$ is just $a(N+1)-a(N-1)$ for a different sequence and so all the inequalities of the lemma follow inductively from the trivial inequality $1>a(2)$.
Lemma 2 
The terms $a(N)$ increase and decrease alternately.
Proof $$a(N+2)-a(N+1)=u_1^{b(N+2)}-u_1^{b(N+1)}.$$ The proof now proceeds similarly to that of Lemma 1. 
Theorem
The $a(2N)$ terms are m.d. to a limit $L$ and the $a(2N+1)$ terms are m.i. to a limit $l$, where $L\ge l$.
Proof 
This is an immediate consequence of Lemmas 1 and 2 and the fact that the terms $a(N)$ are bounded by $0$ and $1$.  
A: Too long for a comment
The general idea is to interpolate the terms to get a function and then analyze its properties.

Let $\{a_n(x)\}$ be a sequence of once-differentiable functions. 
Define the recurrence relation $$A_n(x)=a_n(x)^{A_{n+1}(x)}$$ (where often the '$(x)$' part will be omitted for simplicity.)
Then, we have $$A_n'=A_n\left(A'_{n+1}\ln a_n+A_{n+1}\frac{a_n'}{a_n}\right)$$

Let $$t_n=\frac1n-\frac1{n+1}$$
Let $$H(x)=
\begin{cases}
1, &x<0 \\
\frac{\cos(\pi x)+1}2, &0\le x\le1\\
0, &x>0
\end{cases}
$$
Define $$a_n(x)=(t_n)^{H(n-x)}$$
OP’s sequence  thus becomes
$$\{A_1(1),A_1(2),A_1(3),\cdots\}$$
Then, the limit of the OP's sequence (i.e. $\lim_{n\to\infty}a_{n}$, not to be confused with the $a_n(x)$ in this answer) is $$A_1(\infty)\equiv \lim_{x\to\infty}A_1(x)$$
So our question would become

Does $\lim_{x\to\infty}A_1(x)$ exists?

Let's analyze the derivatives.
Firstly, $$a_n'=-\ln(t_n)H'(n-x)a_n$$
So, $$A_n'=\overbrace{\cdots}^{\text{messy algebra}}=A_nb_n(A_{n+1}H'(n-x)-A'_{n+1}H(n-x))$$ where $b_n=\ln(n(n+1))$.
For $n<\lfloor x\rfloor$, $H'(n-x)=0$. Therefore, we can recursively write out
$$A_1'=\left(\prod^{\lfloor x\rfloor}_{k=1}(-A_kb_k)\right) A'_{\lfloor x\rfloor+1}$$
With $$A'_{\lfloor x\rfloor+1}=A_{\lfloor x\rfloor+1}b_{\lfloor x\rfloor+1}(A_{\lfloor x\rfloor+2}H'(\lfloor x\rfloor+1-x)-\underbrace{A'_{\lfloor x\rfloor+2}H(\lfloor x\rfloor+1-x)}_{=0})$$
we can finally write out something neater
$$A_1'=-\left(A_{\lfloor x\rfloor+2}\prod^{\lfloor x\rfloor+1}_{k=1}(-A_kb_k)\right)\frac{\sin\pi(x-\lfloor x\rfloor)}2$$
We can easily see the alternation of sign in $A_1’$: whenever $x$ increases one, $A_1’(x)$ changes sign. If the product does not converge to zero, then $A_1’(\infty)\ne0$; and, due to the keep changing of sign, one can expect $A_1(x)$ to keep going up and down as $x$ gets larger and larger. Thus one can argue that the limit $A_1(\infty)$ does not exist.
However, I cannot prove the product does not converge to zero.
A: To determine the upper limit (the lower limit can be found similarly)
Let $a_n$ and $u_n$ be as defined in my earlier answer and define $F_n(x)$ to be $${u_{2n-1}^{{u_{2n}}^x}}$$
The following technical result will greatly simplify the working later.
Lemma 3
If $x\ge0.8$ and $u_{2n}\le u_{2n-1}\le0.033$, then $F_n(x)\ge0.8.$ 
Proof
As in Lemma 1, $F_n(x)$ will be minimised (for fixed $u_{2n-1}$) when $x$ is minimised and $u_{2n}$ is maximised.The result therefore follows from the inequality $${t^{t^{0.8}}}\ge 0.8$$ for $t\le 0.033.$
The Inverse function
Since  $F_n(x)$ is m.i. on $[0,1]$, it has an inverse. This is given by
$$G_n(x)=\frac{\ln(\frac{\ln x}{\ln (u_{2n-1})})}{\ln(u_{2n})}.$$
Then $$a_{2n}=F_1(F_2(...(F_n(1))),1=G_n(...(G_2(G_1(a_{2n})))$$
Theorem
The limit $L$ for the @zeraoulia rafik sequence satisfies $$0.8588<L<0.8589.$$
Proof
Direct calculation of the $a_n$ shows that $L<0.8589.$ 
Suppose that  $L\le 0.8588.$ 
Another direct calculation shows that $G_7(...(G_2(G_1(0.8588)))<0.8$ and therefore $G_7(...(G_2(G_1(L)))<0.8$.
Lemma 3 applies to the $G_i$ for $i>7$ and so, as $n\to \infty, G_n(...(G_2(G_1(L)))$  does not tend to $1$, a contradiction.
The calculation given in the theorem can, of course, be carried out to whatever degree of precision is required but I have no reason to question the answer provided by @heropup - I am just giving a proof of the previously obtained numerical result with a method which can be used for other tower sequences and can also be adapted to find lower limits. 
A: $\forall n\in N^{*}:u_{n}=(1-\frac{1}{2})^{(\frac{1}{2}-\frac{1}{3})^{(\frac{1}{3}-\frac{1}{4})^{...(\frac{1}{n}-\frac{1}{n+1})}}}\gt 0$
$u_{1}=v_{1}=1-\frac{1}{2}=\frac{1}{2}$
$u_{2}=v_{1}^{v_{2}},u_{3}=v_{1}^{v_{2}^{v_{3}}},...,u_{n}=v_{1}^{v_{2}^{v_{3}^{....v_{n}}}},v_{n}=\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$
$\ln u_{n}=v_{2}^{v_{3}^{^{....v_{n}}}}\ln v_{1}=-v_{2}^{v_{3}^{^{....v_{n}}}}\ln 2\lt 0$
$0\lt u_{n}\lt 1$
$\frac{1}{9900}^{\frac{1}{10100}}\approx 0.9991$
$\lim_{n \to \infty }(\frac{1}{(n-1)n})^{\frac{1}{n(n+1)}}=1 $
$n\ge 99\Rightarrow v_{n}^{v_{n+1}}\gt  0.999$
$ A_{n}=v_{99}^{v100^{v_{101}^{...v_{n}}}}\longrightarrow \lim_{n \to \infty }A_{n}\approx 1$
$\lim_{n \to \infty }u_{n}=v_{1}^{v_{2}^{^{...v_{99}}}} $
