I try to solve this problem: "Show that if $(M,d)$ is a complete metric space and $f: M \longrightarrow M$ is a map such that exists $p \in \mathbb{N}$ for which $f^p$ is a contraction, then, $f$ have a unique fixed-point $a$ and $\lim_{n\rightarrow \infty} f^n(x) = a$ for any $x \in M$."
My solution: How $f^p$ is a contraction, exists a unique $a \in M$ such that $f^p(a) = a$. Then
$f(a) = f(f^p(a)) = f^p(f(a))$
showing that $f(a)$ is a fixed-point of $f^p$, where $f(a)=a$. Therefore, $a$ is unique because, if $b$ is a fixed-point of $f$, then is a fixed-point of $f^p$.
I'm on trouble in a proof of $\lim_{n\rightarrow \infty} f^n(x) = a$ for any $x \in M$, because $f$ is not-necessary contraction (if $f$ is a contraction, $x_n = f(x_{n-1})$ converges to the fixed-point for any $x_0 \in M$).
First note that in general, $\lim_{n\to\infty} f^n(x)=a$ if and only if for each $s=0,1\dots,p-1$ we have $$\lim_{m\to\infty}f^{mp+s}(x)=\lim_{m\to\infty}(f^p)^m(f^s(x)) =a.\tag{1}$$
To see this, one direction is easy, since if a sequence converges then every subsequence converges to the same limit (we actually don't use this direction but I thought I'd point it out).
For the other direction, if (1) is satisfied then for each $\epsilon>0$ and each $s$ we have some $N_{s,\epsilon}$ so that for $m>N_{s,\epsilon}$ we have $d(f^{mp+s}(x),a)<\epsilon$. Then take $$N_{\epsilon}:=\max_{s\in\{0,1,\dots,p-1\}} (N_{s,\epsilon}p+s),$$ and observe that for any $n>N_\epsilon$, $n$ has the form $n=mp+s$ for some $m>N_{s,\epsilon}$, so $d(f^n(x),a)<\epsilon$.
Finally, observe that the second equation in (1) must hold by the contraction mapping theorem for $f^p$.