So I just checked every number of the form $2^{2^x}$ up to $2^{8192}$ and they all end in $6$. Can someone formally prove that this will be true for all $x$?
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For x a positive integer? – user247327 Mar 24 '16 at 19:09
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Turns out that 2^(2^20) is 23653 characters too long to fit in this comment. – Mathime Mar 25 '16 at 13:59
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The last digit of $2^n$ has a pattern: $2 \to 4 \to 8 \to 6$, so $2^n$ ends in $6$ when $n \equiv 0 \mod 4$.
Since $n = 2^x$, $n \equiv 0 \mod 4$ for $x \geq 2$.
Henricus V.
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Your numbers begin: $2^4, 2^8, 2^{16} \ldots$
Note that each of these can be written as a product of $2^4$s; i.e., of $16$s.
When you multiply numbers whose units digit is a $6$, their product also has a units digit of $6$.
Your numbers are all powers of $16$; so, yes, they will all end in $6$.
Benjamin Dickman
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