It is known that $$W'(z)=\frac{W(z)}{z(1+W(z))}.$$ Let $z=\log t$. Then $x=W(z)$ is a root of the equation $x^x=t$, in particular $x\log x=\log t=z.$ It means that $$W'(z)=\frac{x}{(x+1)\log t}=\frac{1}{(x+1)\log x}=\frac{1}{\log x^{x+1}},\quad x^{x+1}=e^{\frac{1}{W'(z)}}.$$ So solution of the equation $x^{x+1}=a$ is $x=W(z)$, where $z$ is defined by $W'(z)=1/\log a.$

Sorry, my answer is wrong. As it was pointed out by "Simply Beautiful Art" $x\ne W(z)$ but $x=e^{W(z)}.$

It is known that $$W'(z)=\frac{W(z)}{z(1+W(z))}.$$ Let $z=\log t$. Then $x=W(z)$ is a root of the equation $x^x=t$, in particular $x\log x=\log t=z.$ It means that $$W'(z)=\frac{x}{(x+1)\log t}=\frac{1}{(x+1)\log x}=\frac{1}{\log x^{x+1}},\quad x^{x+1}=e^{\frac{1}{W'(z)}}.$$ So solution of the equation $x^{x+1}=a$ is $x=W(z)$, where $z$ is defined by $W'(z)=1/\log a.$