I know that if I have had $y = x^{x+0} $ aka $y = x^x$ I could do
$y = x^x$ // $x = e^{\ln(x)}$
$y=x^{e^{\ln(x)}}$ // $\ln$()
$\ln(y) = \ln(x)e^{\ln(x)}$ then using Lambert's W function I could directly solve for $x$.
$x=e^{W(\ln(y))}$
What do I do in the case I've stated in the title above where $x^x$ becomes $x^{(x+1)}$ I can't directly use the product log since the $x$ in the exponential function and the $x$ in the log function are not the same, or am I wrong?
If you allow, you can solve for $x$ through Lagrange Inversion Theorem.
$$y=x^{x+1}$$
Invert it...
I now have time to work the problem out.
$$f(x)=x^{x+1}$$
$$f(1)=1,f'(1)\ne0$$
$$f^{-1}(x)=1+\sum_{n=1}^{\infty}\lim_{w\to1}\frac{(x-1)^n}{n!}\frac{d^{n-1}}{dw^{n-1}}\left(\frac{w-1}{w^{w+1}-1}\right)^n$$
This is not reducible as far as I know, so you are stuck with this.
If it doesn't converge for the $x$ value you want to calculate, choose a different $a$ value such that $f'(a)\ne0$.
You also need a background understanding of calculus.