ILL 1985 Problem 67

Let $k \geq 2$ and $n_1, n_2, . . . , n_k \geq 1$ natural numbers having the property $n_2 | 2^{n_1} - 1, n_3 | 2^{n_2} -1 , \cdots, n_k | 2^{n_k-1}-1$ , and $n_1 | 2^{n_k} - 1$ . Show that $n_1 = n_2 = \cdots = n_k = 1.$