If $x^{x^{x+1}}=\sqrt{2}$, then evaluate $x^{x^{p}}$, where $p = 2x^{x+1}+x+1$ I can't figure out how to give a proper form to this expression to use the root of two.

If
$$x^{x^{x+1}}=\sqrt{2}$$
find the value of $W$ if
$$W=x^{x^{p}} \quad\text{where}\; p = 2x^{x+1}+x+1$$

EDIT: This is an algebraic manipulation problem with the exponents. There was an error in the previous version (see the Edit History) that I have corrected.
 A: Assuming $x^{x^{x+1}} = \sqrt 2$,
\begin{align*}
x^{x^{2x^{x+1} + x + 1}} &= x^{x^{2x^{x+1}} \cdot x^{x+1}} \\
&= x^{\left(x^{x^{x+1}}\right)^{2} \cdot x^{x+1}} \\
&= x^{\left(\sqrt{2}\right)^{2} \cdot x^{x+1}} \\
&= x^{2 \cdot x^{x+1}} \\
&= \left(x^{x^{x+1}}\right)^{2} \\
&= \left(\sqrt{2}\right)^{2} \\
&= 2
\end{align*}
A: Doing it numerically I get $W=2.7564025221095 $.
Is this what you are after or do you need to represent this with $\sqrt 2$?
A: $x^{x^x} = \sqrt{2} \implies x^x = log_x\sqrt2$
Now:
$W = x^{x^{2x^{x+1}+x+1}}  = x^{\left(x^{2x^{x+1}}x^xx\right)} = x^A$
Let the exponent be A for the time being:
$A = x^{2x^{x+1}}x^xx  =  \left(x^{x^{x+1}}\right)^2x^xx = \left(x^{x^xx}\right)^2x^xx$
Since $x^x = log_x\sqrt2$, we have $A = \left(x^{xlog_x\sqrt{2}}\right)^2xlog_x\sqrt{2} = (\sqrt{2}^x)^2xlog_x\sqrt{2}$
Let $\sqrt{2}^x = B \implies A = B^2log_x(B) = log_x(B^{B^2})$
Remember $W = x^A = B^{B^2} = (\sqrt{2}^x)^{2^x}$
What to do from here?
A: To avoid 3-story exponents, let $y = x^{x+1}$
Then it is given that $x^y = \sqrt{2}$  or $x^{2y} = 2$
$$p = 2y + (x+1)$$
$$x^p = x^{2y} x^{x+1} = 2y$$
$$x^{x^{p}} = x^{2y} = 2$$
