If one $n_i$ is $1$, then all have to be $1$. So lets assume that all are $\neq 1$. Let $i \in \{1,2,...,k\}$ always and define $n_{k+1} :=n_1$ (indices are seen $\mod k$). Let $p_i$ be the smallest prime divisor of $n_i>1$ and let $o_i$ be the order of $2 \mod p_i$ and $q_i$ be the smallest prime factor of $o_i$. We have $o_i|n_{i+1}$ by standard facts on orders, giving $p_{i+1} \leq q_i$. We also have $o_i|p_i-1$, thus $q_i \leq p_i-1$. As a result, $p_{i+1} \leq q_i \leq p_i-1 < p_i$. But this gives $p_1 < p_2 < ... < p_k <p_1$, contradiction.

Here is an other method. $ x = lcm(n_1,..,n_k)$ So $ n_i|2^x - 1,\forall i = 1,..,k$ Imply that $ x|2^x - 1$ So $ x = 1$ (tradition!) http://www.mathlinks.ro/Forum/viewtopic.php?p=849217#849217

See also http://www.artofproblemsolving.com/Forum/viewtopic.php?p=420694#p420694

TTsphn wrote: Here is an other method. $ x = lcm(n_1,..,n_k)$ So $ n_i|2^x - 1,\forall i = 1,..,k$ Imply that $ x|2^x - 1$ So $ x = 1$ (tradition!) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=849217#849217 but this only imply $ x|2^{x}-1 $...can you explain how did you obtain $ n_{i}|2^{x}-1 $?

It has been denoted $ x = \textrm{lcm}(n_1,\ldots,n_k)$. It follows $ n_i \mid 2^{n_{i-1}} - 1 \mid 2^x - 1$, for all $i = 1,\ldots,k$, where $n_0 := n_k$ (indices are taken modulo $k$), since if $a \mid b$, then $2^a - 1 \mid 2^b - 1$. Therefore $ x \mid 2^x - 1$. So $ x = 1$ (by a well known result, whose proof also uses the least prime dividing $x$, as some method above).

Simply consider the smallest prime $p$ that divides one of the $n_i$. Then we get a smallest prime, leading to a contradiction. Thus there is no prime $p$ that divides any of the $n_i$ giving the desired conclusion.

Let the solution least solution be $(n_{1},n_{2},...,n_{k})$ for some $k$.Let $d_{i}$ be the order of $2$ modulo $n_{i}$.Now $(d_{1},d_{2},...,d_{k})$ will be a smaller solution of unless $n_{2}$ $n_{2}$ is the order of $2$ modulo of $n_{1}$ and like that.Therefore $n_{1}$ divides $\phi(n_{2})$.Therefore $n_{1}$ is less than or equal to $n_{2}$.Similarly $n_{2}$ is less than or equal to $n_{3}$ ,.....,$n_{k}$ is less than or equal to $n_{1}$.Therefore all the numbers must be equal and now $n$ divides $2^{n}-1$ only when $n=1$.

Here I post another slightly different solution to this beautiful problem. Peter wrote: Suppose that $k \ge 2$ and $n_{1}, n_{2}, \cdots, n_{k}\ge 1$ be natural numbers having the property \[n_{2}\; \vert \; 2^{n_{1}}-1, n_{3}\; \vert \; 2^{n_{2}}-1, \cdots, n_{k}\; \vert \; 2^{n_{k-1}}-1, n_{1}\; \vert \; 2^{n_{k}}-1.\]Show that $n_{1}=n_{2}=\cdots=n_{k}=1$. Consider the following set $$S=\{(n_1,n_2,\ldots,n_k): n_i\ne 1 \text{for any}~ 1\le i\le k ~\text{and}~ n_{i+1}\mid 2^{n_i}-1~\text{for every i}, \text{where all the indices are taken modulo k}\}$$ Now consider the sum $(n_1+n_2+\ldots+n_k)$. Choose the $n$-tuple from $S$ which has the least sum. For every $i$, we know that $$n_i\mid 2^{\phi(n_i)}-1$$Hence we have that $$n_i\mid 2^{d_{i-1}}-1$$where $d_{i-1}=\gcd(n_{i-1},\phi(n_i))$. But we can also say that $d_{i+1}\mid 2^{d_i}-1$. Now if $d_i<n_{i-1}$ for some $i$, then we can do the following transformation $$(n_1,n_2,\ldots,n_k)\longrightarrow (n_1,n_2,\ldots,n_{i-1},d_i,n_{i+1},\ldots,n_k)$$But note that the new tuple has sum smaller than the "mother tuple", a contradiction. Otherwise if $d_{i}=n_{i-1}$ for all $i$. Then we have $n_i\mid \phi(n_{i+1})\Longrightarrow n_i\le \phi(n_{i+1}) < n_{i+1}$. A contradiction!$~\blacksquare$

if one of $n_i$ s happen to be $1$ then all of them will equal to one. now suppose $p$ is the smallest prime factor of $\prod n_i $ and $p|n_k$ and because $n_k | 2^{n_{k-1}} -1 \implies p|2^{n_{k-1}} -1 \implies 2^{n_{k-1}} \equiv 1 \mod p$ and now as $n_{k-1}$ doesn't have any prime factor less that $p$ so it's co-prime with $p-1$ and because $2^{p-1} \equiv 1 \mod p$ then $p$ must be $1$ which a contradiction which means we can't chose smallest prime factor of $\prod n_i \implies \prod n_i =1$

Say $n_m=1$ then it follows that $n_{m+1}=n_{m+2}=\dots=n_k=n_1=1$ and so $n_1=n_2=\dots=n_{m-1}=1$ too. Now suppose none of them is $1$. Let $p_i$ be the smallest prime divisor of $n_i$. Then $$p_i \mid 2^{n_{i-1}}-1 \implies \text{ord}_{p_i} (2) \mid n_{i-1},$$so $$p_{i-1} < \text{ord}_{p_i} (2) < p_i \implies p_1<p_2<\dots<p_k<p_1,$$a contradiction.