Hint $\ \ $ Prime $\rm\ P\mid N\ \Rightarrow\ 2^{P-1}\equiv 1\equiv 2^N\ \Rightarrow\ 2^{(P-1,\,N)} \equiv 1\ \ (mod\ P)\ $
But if $\rm\:P\:$ is the least prime factor of $\rm\:N\:$ then $\rm\ (P\!-\!1,N) = 1.$
I.e. $\rm\ mod\ P\ $ the order of $\rm\:2\:$ is $\:1\:$ since it divides the coprime integers $\rm\:P\!-\!1\:$ and $\rm\:N.$
Note $\ $ This problem was posted to sci.math on 2009\11\03. $\ $ There I remarked that the proof shows that if $\rm\ A^N\! = 1,\ A\neq 1\,$ then the order of $\rm\:A\:$ is $\,\ge\,$ the least prime factor $\rm\:P\:$ of $\rm\:N.\,$ In particular this implies that $\rm\ A^{P-1}\!\ne 1,\, $ which settles the problem at hand.
For completeness here is a proof of a slightly more general result.
Theorem $\,\ \ \ m,n>1,\,\ m\mid 2^{\large n}-1\,\Rightarrow\, \ell(m) > \ell(n),\ \ $ $\ell(x) =\,$ least prime factor of $\,x $
Proof $\quad\ \, {\rm mod}\,\ q = \ell(m)\!:\,\ 2^{\large n}\equiv 1\equiv\, 2^{\large\color{}{q-1}}\,$ $\Rightarrow$ $\,\ \ell(m) = q > {\rm ord}\,2\color{#80f}{ \ge \ell(n)}$
Remark $\,\ $ The $ $ key idea $ $ is: $\ \ 2^n\equiv 1,\,\ \color{#0a0}{2\not\equiv 1}\,$ implies the order of $\,2\,$ is $\,\color{#80f}{\ge\ {\rm least\ prime}\,\ p\mid n},\ $ because the order must divide $\,n,\,$ and is $\color{}{\neq \color{#c00}1}$ (else $\,\color{#0a0}{2^{\color{#c00}{\large 1}}\equiv 1}),\,$ and the least divisor $>1$ of $\,n\,$ is its least prime factor (by existence and uniqueness of prime factorizations).