Behavior of integer recurrence $u_0\geq 2$, $u_{n+1}=d(u_n)^\alpha$, for $\alpha \geq 3$, where $d$ is the number-of-divisors function Let $u_0\ge 2$ and $\alpha\ge 1$ integers.
I'm trying to study the sequence $(u_n)_{n\ge 0}$ defined by :
$$\forall n\ge 0,\quad u_{n+1}=d(u_n)^{\alpha}, $$
where $d$ is the number-of-divisors function.
If $\alpha=1$, it's not hard to prove that $$\lim\limits_{n\rightarrow +\infty}u_n=2,$$
And if $\alpha=2$ we also have : $$\lim\limits_{n\rightarrow +\infty} u_n=9. $$
However, if $\alpha\ge 3$, the sequence doesn't always converge.
For example, when $\alpha=3$, if $u_0=2$, then for $n\ge 1$, $u_{2n}=2^6$ and $u_{2n+1}=7^3$.

Do we have more results about the behaviour of the sequence $(u_n)_{n\ge 0}$ when $\alpha\ge 3$ ?

$\textit{Edit 2 -}$ After many simulations, I have the following conjecture :
$\boxed{\textbf{Conjecture -} \text{ For all $\alpha\ge 1$, the sequence $(u_n)_{n\ge 0}$ is eventually periodic.}}$
When $\alpha=4$, the sequence always seems to converge, but the limit depends on $u_0$ (for $u_0=2$ the limit is $625$ and for $u_0=6$ the limit is $6561$).
For example, when $\alpha=5$, if $u_0=2$, then the sequence converges to $3^52^5$, but if $u_0=2^53^511^{10}$, then the sequence is $2$-periodic (with $u_1=2^{10}3^{10}11^{5}$).
We can easily prove that :
$\boxed{\text{For all $\alpha\ge 1$, there are infinitelty many numbers such that $(u_n)_{n\ge 0}$ converges.}} $
Indeed, if $u_0=p_1...p_l$ is a squarefree integer, then : $$u_1=2^{\alpha l} $$ and $$u_2=(\alpha l+1)^{\alpha}. $$ By Dirichlet's theorem on arithmetic progressions, there are infinitely many integers $l\ge 1$ such that $\alpha l+1$ is a prime number. 
If we choose a such integer $l\ge 1$, then for all $n\ge 2$ : $$u_n=(\alpha l+1)^{\alpha}, $$ and the sequence $(u_n)_{n\ge 0}$ converges. $\blacksquare$
 A: I'll show that the conjecture that the sequence will end up periodic (or converge which just means of period $1$).
Let's first express the sequence in terms of $v_n=d(u_n)$ so that $v_{n+1}=d(v_n^\alpha)$.
I'll start off by proving that, for any integer $\alpha>0$, there is at most a finite number of positive integers $v$ for which $d(v^\alpha)\ge v$, and then use this to prove that the sequence must eventually be periodic.
Prime factorise $v=p_1^{m_1}\cdots p_k^{m_k}$. Then,
$d(v^\alpha) = (\alpha m_1+1)\cdots(\alpha m_k+1)$.
So, I wish to prove that for $v$ sufficiently large, $d(v^\alpha)/v<1$. We can write this fraction as
$$
\frac{d(v^\alpha)}{v}=\prod_{i=1}^k \frac{\alpha m_i+1}{p_i^{m_i}}.
$$
For any given prime number $p$, the largest value of $(\alpha m+1)/p^m$ is $(\alpha+1)/p$ if $p\le\alpha$ (corresponding to $m=1$), and $1$ for $p>\alpha$ (corresponding to $m=0$). This gives us an upper bound $d(v^\alpha)/v\le A$ where
$A=\prod_{\text{prime } p\le\alpha} (\alpha+1)/p$. Since picking $m_i$ which maximise $d(v^\alpha)/v$ makes it at most $A$, if there is any term with $(\alpha m_i+1)/p_i^{m_i}<1/A$, that will make $d(v^\alpha)/v<1$. However, for any prime $p$, the restriction that $(\alpha m+1)/p^m\ge 1/A$ places an upper bound $m\le M(p)$, as well as an upper bound $p\le 2A$ om primes for which $m\ge 1$.
So, to summarise, in order to get $d(v^\alpha)/v\ge1$, we must have $v=\prod p_i^{m_i}$ where $p_i$ are prime numbers $\le A/2$ and $m_i\le M(p_i)$. There are only a finite number of such numbers. Thus, we can conclude that for all but a finite number of $v$, we have $d(v^\alpha)<v$. Since there is only a finite number of $v$ with $d(v^\alpha)\ge v$, these must have an upper bound $W\ge d(v^\alpha)\ge v$. We can therefore conclude that the sequence $v_n$ will eventually drop and stay below $W$ which means it must eventually become periodic.
Have corrected the explanation of the upper bound $W$. Hope the rest should be correct.
