OK, we have $a_1=a_1,a_2=a_1+d,a_3=a_1+2d\ldots a_n=a_{n-1}+(n-1)d$. Then since $i\mid a_i=a_1+(i-1)d$ for $1\le i \le n-1$, we have $i|a_1-d$; but $n\nmid a_1-d$. If $n=ab$ for two coprime positive integers $1<a,b<n$. Then $a|a_1-d$ and $b|a_1-d$ which implies $n|a_1=b$, of course a contradiction. Therefore $n$ is a prime power.

My solution is pretty similar to yours, but i liked it too much to not post it. So here it goes: Let $d$ be the common difference of the arithmetic progression then: $a_{i}=a_{1} + (i-1)d$ $\forall i=2,...,n$, or setting $a_{1}=a_{0}+d$ we get that $a_{i}= a_{0}+id$ $\forall i=1,...,n$ from that we get that the conditions are equivalent to $[1,...,n-1]|a_{0}$ and $n\nmid a_{0}$. If $n={p_{1}}^{\alpha_{1}}...{p_{k}}^{\alpha_{k}}$ with $k\geq2$ then ${p_{i}}^{\alpha_{i}}<n$ and so ${p_{i}}^{\alpha_{i}}| [1,...,n-1]$ $\forall i=1,...,k$ and because $({p_{i}}^{\alpha_{i}},{p_{j}}^{\alpha_{j}}) = 1$ for $i\not=j$ then $n={p_{1}}^{\alpha_{1}}...{p_{k}}^{\alpha_{k}}$$|[1,...,n-1]$ and by transitivity $n|a_{0}$ Contradiction. Therefore $n$ must be the power of a prime. An example of such a sequence is $a_{i}=(n-1)! + ik$ for $i=1,...,n$ where $n={p_{1}}^{\alpha_{1}}$ where $p_{1}$ is an arbitrary prime.

my solution: Let $a_1=a,a_2=a+d,a_3=a+2d,\cdots ,a_n=a+(n-1)d$ from the condition $i\mid a+(i-1)d\longrightarrow i\mid a-d$ for every $1\le i\le n-1$. so $\text{lcm} [1,2,\cdots ,n]\mid a-d$ but $n\nmid a-d$ if $n$ is composite number the we can write $n=ts$ where $(t,s)=1$ and $t,s>1$ but since $t,s<n$: $ts\mid [1,2,\cdots ,n]\longrightarrow n\mid a-d$ contradiction so $n$ is prime number. DONE

Assume to the contrary, that $n=xy$ for some coprime $x$ and $y$ less than $n$. From relations $x|a_x$ , $y|a_y$, $x|(n-x)|a_n-a_x$ and $y|(n-y)|a_n-a_y$ it follows that $x|a_n$ and $y|a_n$, so that $n|a_n$, which contradicts the hypothesis. Hence the initial statement was true.