<s>[INST] A number $x$ such that $x^n = x$ for every $n$ Is there proof fully algebraic that shows that $1$ is the only number that when put in the function $f(x) = x^n$, for $n = $ any number, the output is always $x$? 
This is pretty obvious, but you seem to prove it using the trivial knowledge . 
For example, $1^4 = 1, 1^{46373} = 1$, but $2^4 \neq 2$
 [/INST] For positive integer $n$, write
$$f(x)=x^n=x^n-1+1=(1+x+\cdots+x^{n-1})(x-1)+1.$$
If $x > 1$, then $(1+x+\cdots +x^{n-1})(x-1) > n(x-1) $ so $f(x) > n(x+1)+1 > 1$.
If $0<x < 1$, then $(1+x+\cdots +x^{n-1})(x-1) < \frac{1}{1-x}(x-1) = -1$. Hence $f(x) < -1+1 = 0$.
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