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The equation is $x!+(x-1)!+(x-2)!+...+1!+0!=x^x+x^{x-1}+x^{x-2}+...x^1+x^0$

I got a solution (1) by substituting $1$ in $x$
I want to know if there is another way to get the solution and also another solution to this equation.

Shraddheya Shendre
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Du Brisingr Arget
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  • $x$ must be a non-negative integer. or x! is not defined. $x^x$ grows faster than $x!$ so once the series on the right is bigger than the series on the left, you can stop looking. $x=0,1$ are the only solutions. – Doug M Nov 24 '16 at 04:10
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    Not a rigorous proof by any means, but when you look at $x! = x \cdot (x-1) \cdot (x-2) \cdots 2 \cdot 1$, you can see that each term in $x!$ is less than or equal to $x$, which means $x! < x^x$ for all $x \ge 2$. It should be fairly easy to test the few starting cases, because once $\sum x^x$ catches up to $\sum x!$, $\sum x!$ will never catch back up. – 2012ssohn Nov 24 '16 at 04:12

1 Answers1

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If you only care about natural numbers, then:

$x! = x \times (x-1) \times \ldots \times 2 \times 1 < x \times x \times \ldots \times x \times x = x^x$ when $x > 1$. Similarly for all the terms you're adding up, so the only possible solutions are 0 and 1, and 0 only works if you use the commonly accepted convention that $0^0 = 0! = 1$.

ConMan
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