Is it true that this function $f(n)=n^{13}$? Assume strictly monotone increasing function;
   such that $f:N^{+}\to N^{+}$, $h$ for all
  $n\in N^{+}$,
  $$f(f(f(n)))=f(f(n))\cdot f(n)\cdot n^{2015}$$
  
  Prove or disprove:$f(n)=n^{13}$  


Put $n=1,f(1)=m$
$$f(f(m))=mf(m)$$
Put $n=m$,
$$f(f(f(m)))=f(f(m))f(m)m^{2015}\Longrightarrow f(mf(m))=m^{2016}(f(m))^2$$
  What about following?
Let us denote by $f^{[k]}(n)$ the $k$th iterate of $f$. I cannot prove the claim, but I can prove that for all integers $n&gt;1$ we have
$$
\lim_{k\to\infty}\frac{\log f^{[k+1]}(n)}{\log f^{[k]}(n)}=13.
$$
This is some kind of asymptotic evidence in favor of $f(n)=n^{13}$ being the only solution - alas, anything but conclusive.

This is seen as follows. We first prove that for all $k\ge3$ we have
$$
f^{[k]}(n)=f^{[2]}(n)^{A_k} f(n)^{B_k} n^{C_k}\qquad(*)
$$
for the sequence of vectors of positive integer determined by the recurrence relations
$$
\left(\begin{array}{r}
A_2\\B_2\\C_2\end{array}\right)=\left(\begin{array}{r}
1\\0\\0\end{array}\right),\qquad
\left(\begin{array}{r}
A_{k+1}\\B_{k+1}\\C_{k+1}\end{array}\right)=M\left(\begin{array}{r}
A_k\\B_k\\C_k\end{array}\right),
$$
where $M$ is the $3\times3$ matrix
$$
M=\left(\begin{array}{crr}
1&amp;1&amp;0\\1&amp;0&amp;1\\2015&amp;0&amp;0\end{array}\right).
$$
The proof follows from the given functional equation of $f$ by induction on $k$. The case $k=3$ is exactly the functional equation. The inductive step follows from the induction hypothesis by substituting $f(n)$ in place of $n$ and again applying the given functional equation.

The eigenvalues of $M$ are $\lambda_1=13$ and $\lambda_{2,3}=-6+i\sqrt{119}$.
The key is that of these $\lambda_1$ has the largest absolute value. Furthermore, if we write the vector 
$$
(A_2,B_2,C_2)^T=x_1e_1+x_2e_2+x_3e_3
$$
in terms of unit eigenvectors $e_1,e_2,e_3$ belonging to the respective eigenvalues, we see that $x_1\neq0$.

For any $k\ge3$ we then have
$$
(A_k,B_k,C_k)^T=\lambda_1^{k-2}x_1e_1+\lambda_2^{k-2}x_2e_2+\lambda_3^{k-2}x_3e_3.
$$
For very large values of $k$ the first component dominates, and consequently
$$
\lim_{k\to\infty}\frac{A_{k+1}}{A_k}=\lim_{k\to\infty}\frac{B_{k+1}}{B_k}=\lim_{k\to\infty}\frac{C_{k+1}}{C_k}=13.
$$
It takes a while to see these limits if you calculate them. For some numerical support I fired up my Mathematica. The entrywise ratios of $M^{129}$ and $M^{128}$ are all in the interval $(12.9,13.2)$.

The claim follows from this by taking logarithms from $(*)$.



I don't know if this helps. It does seem to me that we should concentrate on large values of $n$ and asymptotics first. If only we could prove that $f$ must be a homomorphism of multiplicative monoids. Then it being strictly increasing would force it to a power function, and we know the exponent. If we know that $f$ is a power function the exponent can be determined without resorting to the above asymptotic gymnastics.
26.01.20

$\color{brown}{\textbf{The task analysis.}}$

From the issue conditions should
$$n\in\mathbb N,\quad f(n)\in \mathbb N,\quad f_2(n)=f(f(n)) \in \mathbb N,\quad f_3(n)=f(f(f(n)))\in \mathbb N,\tag1$$

$$n \le f(n)\le f_2(n)\le f_3(n).\tag2$$

Taking in account the issue equation, should
$$n^{2015}\,|\,f_3(n),\quad f(n)\,|\,f_3(n),\quad f_2(n)\,|\,f_3(n).\tag3$$

The issue equation can be presented in the form of
$$\dfrac{f_3(n)}{n^{2197}} = \dfrac{f_2(n)}{n^{169}}\, \dfrac{f(n)}{n^{13}}.\tag4$$

If $\ \exists(N)\forall(n\ge N):\ f(n) &gt; n^{13},$ then
$\ \exists(N)\forall(n\ge N):$
\begin{align}
&amp;\dfrac{f_2(n)}{n^{169}}\, \dfrac{f(n)}{n^{13}} = \dfrac{f(f_2(n))}{n^{2197}} 
&gt; \left(\dfrac{f_2(n)}{n^{169}}\right)^{13}, \\[4pt]
&amp;\dfrac{f(n)}{n^{13}} &gt; \left(\dfrac{f_2(n)}{n^{169}}\right)^{12}
=\left(\dfrac{f(f(n))}{n^{169}}\right)^{12}
&gt; \left(\dfrac{f(n)}{n^{13}}\right)^{156},\\[4pt]
&amp;f(n) &lt; n^{13},
\end{align}

and this contradiction means that 
$$\forall(N)\exists(n\in[N,f_2(N)]):\ f(n) \le n^{13}.\tag5$$

Similarly,
$$\forall(N)\exists(n\in[N,f_2(N)]):\ f(n) \ge n^{13}.\tag6$$

$\color{brown}{\textbf{Solutions building.}}$

From $(4)-(6)$ follows that 
$$f(n) = n^{13}\tag{7}$$
is a solution and asymptotic case for all monotonic solutions.

Since different polynomials can not have infinity set of the common points, $n^{13}$ is the single polynomial, which satisfies $(5-6).$

Other monotonic solutions can be considered as the infinity series of deviations from $(7)$, based on the monotonity conditions in the form of
$$m^{14}\le (m+1)^{13}\tag8$$
or similiar ones.

Analysis shows,that required series are finite.

For example, the function
$$g_{14}(m) = \left\lceil14\,\dfrac{\ln m}{\ln(m+1)}\right\rceil\tag9$$
has the table of the values in the form of



On the other hand, every deviation near $n$ requires corresponding deviation near $f(n),$ and this lead to the secondary monotonity conditions, which can not be provided.

Similar table for the function
$$g_{169}(m) = \left\lceil169\,\dfrac{\ln m}{\ln(m+1)}\right\rceil\tag{10}$$



shows, that solutions are bounded with $m=45$ although $2^{13}=8192.$

Therefore, the solution $(7)$ is unique.