How to solve $\upsilon^\upsilon=\upsilon+1$

I used a very simple method, "iterative fixed point method", look that $x^{x}=x+1$ is equivalent to $x^{x}-x-1=0$ and $-2x=x^{x}-3x-1$ and $x=\frac{x^{x}-3x-1}{-2}$. Define $g(x)=\frac{x^{x}-3x-1}{-2}$ so the answer of equation $x^{x}=x+1$ is the fixed point of the function $g(x)$. For finding it we note that because $g'(x)=\frac{(1+ln(x))x^{x}-3}{-2}$ is decreasing in interval $[1.65,1.9]$ and $g'(1.65)=-0.2124....$ and $g'(1.9)=-0.667....$ so $g'(x)$ is negative in interval $[1.65,1.9]$ and $g(x)$ is decreasing in this interval and as $g(1.65)=1.83....$ and $g(1.9)=1.65....$ we have $$\forall x\in [1.65,1.9] \; :\;g(x)\in [1.65,1.9]$$ and also we have seen that $$\forall x\in [1.65,1.9] \; : \; |g'(x)|\leq 0.67<1$$ so by a theorem in Numerical Analysis for iterative fixed point method the sequence $\{g(x_{n})\}_{n=1}^{\infty}$ is convergence to requested point, if we choose $x_{1}\in [1.65,1.9]$. But if you want something else like only using "Lambert W function" and such things, please say me.