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How to show that $5^{2^{e-2}}$mod$(2^e) = 1$ mod$(2^e)$? ($e \geq 3$)

Thanks

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    **Hint** $\ a\equiv 1\pmod{\!\color{#c00}2n}\,\Rightarrow\, a^2\equiv 1\pmod{\!\color{#c00}4n}\ $ by $\,(1\!+\!2n)^2 = 1\!+\!4n(1\!+\!n)\ \ $ – 2017-02-16
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    Then we have $5^{2^{e-1}} \equiv 1 \mod 2^{e+1}$? And what's the next step? – 2017-02-16
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    That's the inductive step. So it remains to prove the base case $\,e = 3.\ \ $ – 2017-02-16
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    Ooh Yes, I was not thinking of induction, but now I understand it. Thanks!! – 2017-02-16
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    Great. This is a special case of general results sometimes called [LTE = Lifting The Exponent.](http://math.stackexchange.com/a/1996859/242) – 2017-02-16

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Using weak induction,

if $5^n=1+a2^m$ where $a$ is an odd positive integer,

$5^{2n}=(5^n)^2=(1+a2^m)^2=1+2^{n+1}a+a^2(2^n)^2\equiv1\pmod{2^{n+1}}$ for $2n\ge n+1\iff n\ge1$

Now for the base case $5^{2^1}\equiv1\pmod{2^{1+2}}$