A matter of '6'!

Relevant wiki: Finding the Last Digit of a Power

As the cyclicity of powers of $2$ is $2,4,8,6$ , we can have unit digit $6$ in $2^{n!}$ only when $n!$ is divisible by 4. And, this is true for all $n>3$ , because it will have $4$ as a factor.

The last digit of $6^k$ where, $1<=k$ is always 6

$\large 2^{1\times 2\times 3\times \cdots \times n}$ if $n>3$ then,

$\large \begin{aligned} 2^{1\times 2\times 3\times \cdots \times n}&= 2^{1\times 2\times 3\times4\times \cdots \times n}\\ &=(2^{1\times 2\times 3})^{4\times5\times6\times \cdots \times n}\\ &=64^{4\times5\times6\times \cdots \times n}\\ &\equiv 4^{4\times5\times6\times \cdots \times n} \pmod {10}\\ &\equiv (4^2)^{2\times5\times6\times \cdots \times n} \pmod {10}\\ &\equiv 16^{2\times5\times6\times \cdots \times n} \pmod {10}\\ &\equiv 6^{2\times5\times6\times \cdots \times n} \pmod {10} \\ &\equiv 6^k\pmod {10}\\&\equiv 6 \pmod {10}\end{aligned}$

Therefore, it is proved that , for $n>3$ the last digit of $\large 2^{1\times 2\times 3\times \cdots \times n}$ is always 6 .

The last digit of powers of $2^{ }$ can be generalized as

$2^N \pmod{10} = \begin{cases} {\color{#3D99F6}{2^{N \pmod{4}}}} \text{,} & \text{if} \ N \pmod{4} \equiv 1, 2, 3 \\ 2^4 = 16 \equiv {\color{#3D99F6}{6}} \ \text{mod} \ 10 \text{,} & \text{if} \ N \pmod{4} \equiv 0 \end{cases}$

That is, if the power modulus $4^{ }$ is equivalent to $1$ , $2$ , or $3$ , then the last digit is that remainder itself. If the power is congruent to $0 \ \text{mod} \ 4$ - in other words, if $4^{ }$ divides the power - then the last digit is $6$ . This is generalized as such because the powers of $2$ form the periodic sequence $2, 4, 8, 6$ for powers $> 0$ .

Now, observe that $n! \equiv 0 \pmod{4} \ \forall \ n \geq 4$ . This is because all factorials $\geq 4!$ will have a factor of $4^{ }$ .

$\displaystyle \begin{aligned} (n + 1)! & = (n + 1) \cdot n! \ \forall \ n \geq 0 \\ \implies n! & \mid (n + 1)! \ \forall \ n \geq 0 \\ \implies n & \mid (n + 1)! \ \forall \ n \geq 0 & \small \color{#3D99F6} \text{Let} \ n = 4 \\ \implies 4 & \mid (n + 1)! \ \forall \ n \geq 4 \end{aligned}$

Because $n! \equiv 0 \pmod{4} \ \forall \ n \geq 4$ , we can conclude that $2^{n!} \equiv 6 \pmod{10} \ \forall \ n \geq 4$ . $\blacksquare$

In this solution 'number' will mean 'positive integer', and 'power' will mean 'positive integral power'.

First note that if the rightmost digit of a number is 6, then all the powers of that number end in 6. This follows from $6 \times 6 = 36$ and the traditional algorithm for long multiplication. You find a power by multiplying the previous power by your number and start ' six sixes make thirty-six so write down 6 and carry three...'. The six which you write down becomes the rightmost digit of your answer. So if your number has a six in the least significant position, you can never get away from it in the subsequent powers of the number.

Secondly note that $2^{4!}=16777216$ which ends with a six.

Then $2^{5!}=\left(2^{4!}\right)^5$ . Being the power of a number which ends in six, this also ends in six.

Then $2^{6!}=\left(2^{5!}\right)^6$ . Being the power of a number which ends in six, this also ends in six.

and so it goes on and on.

For a gold star you could make this solution more formal by invoking mathematical induction.

If two integers both have a one's digit of 6, their product will also be an integer with a one's digit of 6. Write it out and try it. All you need to do is the first step of the multiplication. 6 * 6 = 36. You would write the 6 down and carry the 3. If you were to complete this multiplication, under that 6 would be a bunch of 0's. And when all is added together at the end, 6 + 0 + 0 +...+ 0 = 6. It will always happen.

And so, mathematical induction starting with $n=4$ :

$2^{4!}=16777216\quad\quad$ Check

A calculator can tell you that, but put that to the power of 5 and a calculator will put the answer in scientific notation and you won't see the last digit. But you don't need to.

Assuming this is true for n, is it true for n+1?

$2^{(n+1)!}=(2^{n!})^{n+1}$

Since $2^{n!}$ is an integer ending in 6, based on our earlier deduction, if this number is multiplied by itself $n+1$ times, the resulting product will also end in 6. $\quad\Box$

Notice that the last digit of $2^{4}$ is $6$ . Raising $2^{4}$ to N where N is a positive integer yields $6$ as the last digit. The requirement for $2^{1 \times{2} \times{3} \times{...} \times{n}}$ or $2^{n!}$ to have 6 as its last digit is that $n!$ must be divisible by $4$ . Values for n greater than 3 or $n > 3$ make $n!$ divisible by 4. Thus, this proves that for any integer $n > 3$ , the last digit of $2^{1 \times{2} \times{3} \times{...} \times{n}}$ is always 6.

$2^{1\times 2\times 3\times 4}\equiv 64^4\equiv 4^4\equiv 256\equiv 6\pmod{10}$

$\underline{\textbf{Proposition:}}$

$6^k\equiv 6\pmod{10}\forall k\in\mathbb{Z^+}$

$\underline{\textbf{Proof:}}$

We proceed by induction:

$\underline{\text{Base Case:}\;n=1}$

In this case we have $6^n\equiv 6^1\equiv 6\pmod{10}$ ,which is true.

$\underline{\text{Inductive Step:}}$

Assume that for some $m$ ,we have $6^m\equiv 6\pmod{10}$ .

Therefore,we have: [6^{m+1}\equiv 6^m\times 6\equiv 6\times 6\equiv 36\equiv 6\pmod{10})

Hence proved.

Therefore,for $n>3$ ,we have: $\color{#D61F06}{\Large{2^{1\times 2\times 3\times 4\times\cdots\times n}\equiv 6^{5\times 6\times\cdots \times n}\equiv 6\pmod{10}}}$

Because $n > 3$ , we have $2^{1 \times 2 \times 3 \times 4 \times \ldots \times n} = (2^4)^{1 \times 2 \times 3 \times \ldots \times n} = 16^{1 \times 2 \times 3 \times \ldots \times n}$ . And $16 \times 16 \times 16 \times ... \times 16$ always ends with digit 6.

$n>3$ , so $n \geq 4$ , and $4 \vert n!$ . Thus, the expression can be written as $(2^2)^{2m}$ , for some integer m. Take this mod 5 to get: $(2^2)^{2m} \equiv (-1)^{2m} \equiv 1 \mod 5$ . The expression must then either end in 1 or 6. Obviously, it is even, so it ends in 6.