If $x\geq 1$, then $x^{x^{\cos(x)}}\geq (1-\sqrt{x}+x)^{(x+1)^{\cos(x)}}$
Reducing to the Case $\boldsymbol{a=1}$
If $a\le1$, then $\left(\frac{x}{x+1}\right)^a\ge\frac{x}{x+1}$. Therefore, since $x\ge1$, we have $$ x^{\left(\frac{x}{x+1}\right)^a}\ge x^{\frac{x}{x+1}}\tag1 $$ Thus, if we can show that for $x\ge1$, $$ x^{\frac{x}{x+1}}\ge x-\sqrt{x}+1\tag2 $$ then inequality $(1)$ says that for $a\le1$, $$ \bbox[5px,border:2px solid #C0A000]{x^{x^a}\ge\left(x-\sqrt{x}+1\right)^{(x+1)^a}}\tag3 $$
Proving the Case $\boldsymbol{a=1}$
Substituting $x\mapsto x^2$ in $(2)$, we see that, for $x\ge1$, showing $(2)$ is equivalent to showing $$ x^2\log\left(x^2\right)\ge\left(1+x^2\right)\log\left(x^2-x+1\right)\tag4 $$ That is, inequality $(4)$ shows inequality $(3)$.
Note that $$ \begin{align} &x^2\log\left(x^2\right)-\left(1+x^2\right)\log\left(x^2-x+1\right)\\ &=x^2\log\left(\frac{x^2}{x^2-x+1}\right)-\log\left(x^2-x+1\right)\\ &=-x^2\log\left(1-\frac{x-1}{x^2}\right)-\log\left(x^2-x+1\right)\\ &\ge x^2\left(\frac{x-1}{x^2}+\frac12\frac{(x-1)^2}{x^4}+\frac13\frac{(x-1)^3}{x^6}+\frac14\frac{(x-1)^4}{x^8}\right)-\log\left(x^2-x+1\right)\\ &=x-\frac12-\frac2{3x}-\frac1{4x^2}+\frac7{6x^4}-\frac1{x^5}+\frac1{4x^6}-\log\left(x^2-x+1\right)\tag5 \end{align} $$ The derivative of the right side of $(5)$ is $$ \begin{align} &1+\frac2{3x^2}+\frac1{2x^3}-\frac{14}{3x^5}+\frac5{x^6}-\frac3{2x^7}-\frac{2x-1}{x^2-x+1}\\ &=\frac{(x-1)^3\left(6x^6-2x^4-x^3+4x^2-12x+9\right)}{6x^7\left(x^2-x+1\right)}\tag6 \end{align} $$ For $x\ge1$, $$ \begin{align} &6x^6-2x^4-x^3+4x^2-12x+9\\ &=6(x-1)^6+36(x-1)^5+88(x-1)^4+111(x-1)^3\\ &\phantom{=}\ +79(x-1)^2+21(x-1)+4\\ &\ge4\tag7 \end{align} $$ Therefore, $(6)$ and $(7)$ show that the right hand side of $(5)$ is increasing for $x\ge1$. This verifies $(4)$ since the right hand side of $(5)$ is $0$ for $x=1$. Thus, we have proven $(3)$.
$\Large\square$
Closeness of the Inequality
Consider the Taylor series for $x^2\log\left(x^2\right)-\left(1+x^2\right)\log\left(x^2-x+1\right)$ near $x=1$. Let $x=1+u$, then look at the Taylor series $$ \begin{align} &x^2\log\left(x^2\right)-\left(1+x^2\right)\log\left(x^2-x+1\right)\\ &=2\left(1+2u+u^2\right)\log(1+u)-\left(2+2u+u^2\right)\log\left(1+u+u^2\right)\\ &=2\left(1+2u+u^2\right)\left(u-\frac{u^2}2+\frac{u^3}3-\frac{u^4}4+\frac{u^5}5-\frac{u^6}6+\frac{u^7}7+O\!\left(u^8\right)\right)\\ &\phantom{=}\ -\left(2+2u+u^2\right)\left(u+\frac12u^2-\frac23u^3+\frac14u^4+\frac15u^5-\frac26u^6+\frac{u^7}7+O\!\left(u^8\right)\right)\\ &=\frac{u^4}6-\frac{u^5}6-\frac{u^6}{60}+\frac{u^7}5+O\!\left(u^8\right)\\ &=\frac{(x-1)^4}6-\frac{(x-1)^5}6-\frac{(x-1)^6}{60}+\frac{(x-1)^7}5+O\!\left((x-1)^8\right)\tag8 \end{align} $$ The series in $(8)$ shows why the inequality is very close when $x\sim1$.
The asymptotic series in $(5)$ shows that as $x\to\infty$, we have $$ x^2\log\left(x^2\right)-\left(1+x^2\right)\log\left(x^2-x+1\right)=x-2\log(x)-\frac12+O\!\left(\frac1x\right)\tag9 $$ which shows that as $x\to\infty$, $$ \frac{x^x}{\left(x-\sqrt{x}+1\right)^{x+1}}=\frac{e^{\sqrt{x}}}{x\sqrt{e}}\left(1+O\!\left(\frac1{\sqrt{x}}\right)\right)\tag{10} $$