An indices problem

What is the biggest x such that there is a number n such that x^{n}>n?

I have a rough idea of what the answer is, but I got it by guess-and-check. How do i solve the question (apart from guess-and-check)? I think that x should be somewhere near 1.444667732455312658679247306 but why?

I think e could be the n such that x^{n}-n is greatest.

Please answer if you have an idea of how to solve this.

#NumberTheory

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$