Given your original function, $f(x) = x^{x+1} - (x+1)^x = 0$, you can define the Taylor Series expansion around some value v to give an expansion such that f( x) = a for some a, which in itself isn't helpful. However, there exists an inversion, using the Lagrange Inversion Theorem, which allows you to specify the a which your f( x) will return and gives you the corresponding x.
The first few terms of the series are:
$a+frac{x-f[a]}{f'[a]}-frac{f''[a] (x-f[a])^2}{2 f'[a]^3}+frac{left(3 f''[a]^2-f'[a] f^{(3)}[a]right) (x-f[a])^3}{6 f'[a]^5}+frac{left(-15 f''[a]^3+10 f'[a] f''[a] f^{(3)}[a]-f'[a]^2 f^{(4)}[a]right) (x-f[a])^4}{24 f'[a]^7}$
Remember, though: this is HIGHLY volatile when you're not near the actual value. Using 3 makes the value explode upwards, and using 2 makes it explode in the negative direction. However, a value of 2.2 will converge to the value you're looking for. You can use the Inversion Theorem to calculate more terms in the inverse series and use it to calculate your constant to arbitrary precision, given enough computational time.
Of course, this isn't what you're looking for (i.e. closed form or 'nice' series), but it's the best that I've found for this particular problem. Unfortunately, it converges so slowly with respect to computational time that it probably becomes more convenient for you to use the secant method to get any sort of good approximation.
--Gabriel Benamy
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