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Show that by induction method that $2^{2^n}+1$ has $7$ in unit's place for all $n\geq 2$.

I have tried to show this with the following way :

Let $f(n)=2^{2^n}+1$.
Then for $n=2,f(2)=2^{2^2}+1=17\Rightarrow f(2)\equiv 7(\mod 10) $
Suppose for $n=m$, the result is true i.e., $f(m)=2^{2^m}+1=10p+7$, where $p$ is an integer.
How can I show this result for $n=m+1$ ?

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$f(1)=5$; you want $f(2)$ in that base case calculation. – Brian M. Scott Aug 15 '13 at 17:16
You’re making much too big a deal of this, I believe. Did you actually evaluate $2^{2^2}$ and $2^{2^3}$? – Lubin Aug 15 '13 at 17:18
Sorry for my mistake. A correction has been done. – Md Saiful Gazi Aug 15 '13 at 17:19

HINT: $$f(m+1)=2^{2^{m+1}}+1=2^{2^m\cdot2}+1=\left(2^{2^m}\right)^2+1=\big(f(m)-1\big)^2+1$$

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Hint: Note that $2^{2^{n+1}}=\left(2^{2^n}\right)^2$ and $(10k+6)^2=100k^2+120 k+36$

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HINT: When you square a number ending in $6$, don’t you get another such number?

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Hint: apply induction to $2^{2^m}$

Additional possibility: investigate the units digits of small powers of $2$ and formulate and prove a stronger (but simpler) hypothesis by induction of which the given statement is a special case.

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Let the statement F(n) be true and with F(n+1) the power of 2 that is 2^m becomes 2^(n+1) which is a number of the form 4k. And 2^4k will always yield a 6 in its last digit and by adding 1 you will get 7 in the last digit.

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