Does $n | ϕ (2^{n} - 1)$ ?

By Graeme Mcrae on Monday, July 22, 2002 - 08:24 am:

Is it true that Euler's Totient $n | ϕ (2^{n} - 1)$ ?
Can it be proven?

First few examples:

n	2ⁿ -1	phi(2ⁿ -1)	phi/n
1	1	1	1
2	3	2	1
3	7	6	2
4	15	8	2
5	31	30	6
6	63	36	6
7	127	126	18
8	255	128	16
9	511	432	48
10	1023	620	62
11	2047	1936	176
12	4095	1728	144
13	8191	8190	630

References:

reference 1

reference 2

So is it true for all n?

Can it be proven? I would like to see a proof.

[Editor: Yatir and David came up with a proof which uses the following steps:
(1) The order of 2 modulo 2ⁿ -1 is n.
(2) The group of invertible elements modulo 2ⁿ -1 has order phi ((2ⁿ )-1).
(3) Since the order of an element in a group divides the order of the group, combine (1) and (2) to obtain the result.]