Call a sequence $L=\left\{l_n\right\}_{n\in\mathbb Z_{>0}}$ of nonnegative integers logoana if $k\mid l_k$ and $l_{n!}=\sum_{k=1}^n l_n$. Suppose that $L$ is not the zero sequence. It follows that for all sufficiently large $n$ we have $l_{n!}=\sum_{k=1}^n l_k>0$. Observe also that $l_k=l_{k!}-l_{(k-1)!}$ for all $k\in\mathbb Z_{>0}$. Repeatedly expanding, $$l_{n!}=l_{(n!)!}-l_{(n!-1)!}=l_{((n!)!)!}-l_{((n!)!-1)!}+l_{(((n!)-1)!)!}-l_{(((n!)-1)!-1)!}=\text{ etc..}$$From which it follows that $T^{i}(n!)\mid l_{n!}\implies l_{n!}>T^i(n!)$ where $T(x)=(x-1)!$ and $i$ is arbitrarily large, a contradiction for all $n>2$. Now note that for any alagoana sequence $A=\left\{a_n\right\}_{n\in\mathbb Z_{>0}}$ and prime $p$, the sequence $\left\{v_p(a_n)\right\}_{n\in\mathbb Z_{>0}}$ is logoana, and thus that $v_p(a_n)=0$ for all primes $p$ and positive integers $n$, implying that $A=\boxed{\left\{1\right\}_{n\in\mathbb Z_{>0}}}$ is the only alagoana sequence $\blacksquare$

For each prime number $p$. Let $\{ b_n\}$ be the sequence of non-negative integer defined by $\nu_p(a_n)=nb_n$. We have $n!b_{n!}=\sum_{i=1}^{n}{ib_i}\implies (n+1)b_{n+1}=n!((n+1)b_{(n+1)!}-b_{n!})\implies \frac{n!}{\gcd (n!,n+1)} \mid b_{n+1}$ for all $n$. Let $b_{n+1}=\frac{n!}{\gcd (n!,n+1)} c_{n+1}$ for all $n$, we get $\frac{n+1}{\gcd (n!, n+1)} c_{n+1}=(n+1)b_{(n+1)!}-b_{n!}$. We have $\frac{(n!-1)!}{n!}=\frac{(n!-1)!}{\gcd ((n!-1)! ,n!)}\mid b_{n!}\implies \frac{((n+1)!-1)!}{n!} \mid (n+1)b_{(n+1)!} $ for all $n>3$. So, $\frac{(n!-1)!}{n!}\mid (n+1)b_{(n+1)!}-b_{n!}\mid (n+1)c_{n+1}\implies \frac{(n!-1)!}{(n+1)!} \mid c_{n+1}$ for all $n>3$. Hence, $\frac{(n!-1)!}{(n+1)!} \mid b_{n+1}$ for all positive integer $n>3$. Let $f_1(n)=\frac{(n!-1)!}{(n+1)!}$ for all $n>3$. In general, suppose we get $f_i(n)\mid b_{n+1}$ for all positive integer $n>3$ for some function $f_i:\mathbb{Z}_{>3}\to \mathbb{Z}^+$ satisfy $(n+2)f_i(n!)\mid f_i((n+1)!)$ for all $n>3$, and $(n+1)\mid f_i(n!)$ for all $n>3$. We get $\frac{f_i(n!)\times n!}{n+1}\mid b_{n+1}$ and also $f_i((n!)!)\mid f_i(((n+1)!-1)!) \mid \frac{f_i(((n+1)!)!)}{(n+1)!+1} \implies (n+2)\frac{f_i((n!)!)\times (n!)!}{n!+1} \mid \frac{f_i(((n+1)!)!)\times ((n+1)!)!}{(n+1)!+1}$ for all $n>3$, and $n+1\mid \frac{f_i((n!)!)\times (n!)!}{n!+1}$ for all $n>3$. So, the function $f_{i+1}:\mathbb{Z}_{>3}\to \mathbb{Z}^+$ defined by $f_{i+1}(n)=\frac{f_i(n!)\times n!}{n+1}$ for all positive integer $n>3$ satisfy $f_{i+1}(n)\mid b_{n+1}$ for all $n>3$ and also the two extra constraints. Moreover, $f_{i+1}(n)> f_i(n)$ for all $n>3$. Hence, by induction, $f_i(n)\mid b_n$ for all positive integer $i$, this forces $b_n=0$ for all $n> 3$, and so for all positive integer $n$. Done.

$a_{n!}=a_1a_2a_2...a_{n-1}a_n$, and $a_{(n-1)!}=a_1a_2a_2...a_{n-1}$ so, $a_{n!}=a_na_{(n-1)!}$. $a_{n!}$ is a $n!-$power, and, in particular, a $(n-1)!-$power. $a_{(n-1)!}$ is also a $(n-1)!-$power, so, we can conclude that $a_n$ is a $(n-1)!-$power. Now, we have that $a_{n!}$ is a $(n!-1)!-$power, and, in particular, a $((n-1)!-1)!-$power. $a_{(n-1)!}$ is also a $((n-1)!-1)!-$power, so, we can conclude that $a_n$ is a $((n-1)!-1)!-$power. Repeat this process a lot of times and you get that $a_n$ is a $((((n-1)!-1)!-1)!...-1)!-$power. It means that, for every $n\geq4$, you can aways increase the power, so, we conclude that $a_n=1$, for every $n\geq4$. $a_6=1 \implies a_1a_2a_3=1 \implies a_1=a_2=a_3=1$. So, the only solution is $a_n=1$.