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.

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*}
