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Any recommendations on how to solve this using a power series:

$$ x^{x+1}=(x+1)^x $$

Figured out you could not do it algebraically so decided to think outside the box.

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    Well, do you think there is closed form or do you just want an approximation? – 2017-02-17
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    I know the solution from graphing is approx 2.29 so I believe there should be a closed form? Not sure so thought it might be easier to use a log property, but not sure how or which series to use and steps I should use to combine them. – 2017-02-17
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    Well, I don't know but [computation](http://www.wolframalpha.com/input/?i=x%5E(x%2B1)%3D(x%2B1)%5Ex) doesn't reveal a possible closed form. – 2017-02-17
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    How close to a close form do you think I could approximate? Just out of curiosity? – 2017-02-17
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    @John It doesn't make sense to approximate a closed form. It either is closed form, or it isn't. – 2017-02-17

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Note that

$$x^{x+1}=x^x\cdot x=(x+1)^x$$

Divide both sides by $x^x$ to get

$$x=\left(1+\frac1x\right)^x$$

Now from here I'll tell you I don't believe there is a closed form, but we can do some quick fixed-point iteration:

$$x_{n+1}=\left(1+\frac1{x_n}\right)^{x_n}$$

With $x_0=2.3$, we get

$x_1=2.2940772541106$

$x_2=2.2932879508444$

$x_3=2.2931825401271$

$x_4=2.2931684586440$

$x_5=2.2931665774723$

Which is the first few digits of the solution.

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    So is it possible to tell what this solution is besides guessing based on initial value? Was wondering if it would be possible to use a Taylor series approximation on a series that converged to determine what the solution is? Since Algebraically this is all that it seems to be simplified down to ? – 2017-02-19
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    Since we know as $x\to\infty$, the right side monotonically approaches $e$ from below, and as $x\to0$, the right side approaches one, so we may, without guessing, deduce that we can have $1$x_0=1.9$ – 2017-02-19
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It's actually not that difficult to get the power series. Using the fact that any number $a^x$ is equal to $e^{x\ln a}$ we can express the power series as $\sum\limits_{n=0}^{\infty}\frac{(x\ln a)^n}{n!}$ from the power series for $e^x$. If a is another variable say $x$ then we just replace it with x. So here we have a power series for both the expressions given