Prove that sequence converges to the non-negative root of $a^p - a = 1$ Let $p > 1$ and $x_n = \sqrt[p]{1 + \sqrt[p]{1 + \sqrt[p]{1 + \dotso + \sqrt[p]{1}}}}$  (n radicals).
Prove that $\{x_n\}$ converges to the positive root of $a^p - a = 1$.
How do I find the roots of such equation or do I even need to find them?
How do I usually deal with problems, when I have a lot of radicals one under each other?
P.S. Ok, I noticed $x_n^p -1 = x_{n-1}$
 A: It's easy to prove by induction that:


*

*$x_n$ is increasing

*$1\le x_n\le 2$ (so $x_n$ is bounded)
So $x_n$ is convergent ($x_n\rightarrow a>0$). By taking the limit in both sides of the original recurrence, one gets $a^p-1=a$
Proof of 2.
$x_1=1\in[1,2]$
Suppose $1\le x_{n-1}\le 2$ Then $1\le 1+x_{n-1}\le 3$
So $1\le \sqrt[p]{1+x_{n-1}}\le \sqrt[p]{3}\le \sqrt 3\le \sqrt 4=2$
which is the same to $1\le x_n\le 2$
Note: I assumed that $p$ is positive integer ($p\ge 2$) because of the root, but it's easy to adapt the above to hold any real $p>1$, by replacing $2$ with $\frac{p}{p-1}$, and using (generalized) Bernoulli's inequality to prove by induction $1\le x_n\le \frac{p}{p-1}$
$x_n=(1+x_{n-1})^\frac{1}{p}\le 1+\frac{x_{n-1}}{p}\le 1+\frac{1}{p-1}=\frac{p}{p-1}$
Proof of 1.
$x_1=1<x_2=\sqrt[p]{2}$.
Suppose $x_{n-1}<x_n$. Then $\sqrt[p]{1+x_{n-1}}<\sqrt[p]{1+x_n}$, which is $x_n<x_{n+1}$
A: Let $\phi(x) = x^p-(1+x)$ and notice that $\phi(0) = -1$, $\phi'(0) =-1$,
$\phi''(x) \ge 0$ and $\lim_{x \to \infty} \phi(x) = \infty$.
In particular, there is a unique $x^*>0$ such that $\phi(x^*) = x^*$, and
for $0 \le x \le x^*$, we have $\phi(x) <0$ and for $x^* \le x$, we have
$\phi(x) >0$.
Since $x \mapsto \sqrt[p]{x}$ is strictly increasing, we see that
$\eta(x) = x - \sqrt[p]{1+x}$ also satisfies the same conditions regarding sign.
Let $f(x) = \sqrt[p]{1+x}$ and note that $f$ is strictly increasing. 
From above we see that there is a unique $x^*>0$ such that
$f(x^*) = x^*$.
If $0\le x \le x^*$
then $x \le f(x) \le f(x^*) = x^*$.
If $x^* \le x$ then $x^* = f(x^*) \le f(x) \le x$.
Now consider the iteration $x_{n+1} = f(x_n)$ starting from any $x_0$.
If $0 \le x_0 \le x^*$, then $x_n$ is a non decreasing sequence that is bounded above, hence $x_n \to \tilde{x}$ and since $f$ is continuous, we
have $\tilde{x} = f(\tilde{x})$ and so $\tilde{x} = x^*$.
A similar analysis shows that if $x^* \le x_0$, then $x_n \to x^*$.
