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 $\infty$  ? 

$\forall n\in N^{*}:u_{n}=(1-\frac{1}{2})^{(\frac{1}{2}-\frac{1}{3})^{(\frac{1}{3}-\frac{1}{4})^{...(\frac{1}{n}-\frac{1}{n+1})}}}\gt 0$
$u_{1}=v_{1}=1-\frac{1}{2}=\frac{1}{2}$
$u_{2}=v_{1}^{v_{2}},u_{3}=v_{1}^{v_{2}^{v_{3}}},...,u_{n}=v_{1}^{v_{2}^{v_{3}^{....v_{n}}}},v_{n}=\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$
$\ln u_{n}=v_{2}^{v_{3}^{^{....v_{n}}}}\ln v_{1}=-v_{2}^{v_{3}^{^{....v_{n}}}}\ln 2\lt 0$
$0\lt u_{n}\lt 1$
$\frac{1}{9900}^{\frac{1}{10100}}\approx 0.9991$
$\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}}}} $
