Last digit

For finding the last digit of the number it is sufficient to calculate its remainder on division by 10

By summing the GP we get $N=2^{2016}-1$

Now $2^{4k} \equiv 6 \pmod{10}$

$\implies 2^{2016} \equiv 6 \pmod{10}$ (since $2016$ is a multiple of $4$ )

$\implies 2^{2016} -1 \equiv 5 \pmod{10}$

$\implies N \equiv 5 \pmod{10}$

$\implies (N+9) \equiv 4 \pmod{10}$

Since $N= 2^{2016}-1$ is odd and $4^{odd} \equiv 4 \pmod{10}$

$\implies (N+9)^N \equiv 4 \pmod{10}$

Hence the last digit is $\boxed{4}$

Since 2 has order of 4 in powers i.e. unit digits repeat after every 4 occurrences => Unit digits which keep on getting repeated are 2, 4, 8, 6 { sum = 20 => unit digit 0 }

Starting from power 1 to 2012 => unit digit => 0 2013, 2014, 2015 result in {2+4+8} => unit digit 4

As 1 is the first term 4+1=5 so unit digit of N is 5 => Unit digit of 9+N => 4 Our answer of unit digit of 4^5 => 4 :) :)

The cyclicity of 2 is 4.

That is $2^{n}$ ends with 2 if n leaves a remainder of 1 when divided by 4.

And ends with 4 if n leaves a remainder of 2 when divided by 4.

And ends with 8 if n leaves a remainder of 3 when divided by 4

And ends with 6 if n is divisible by 4

So last digit of N =(1+(2+4+8+6)53 +2+4+8)%10=((20)53+15)%10

As 20(53) will be divisible by 10 but 5 is not therefore the remainder is 5

=>Unit digit of N is 5

=>Unit digit of $(9+N)^{N}$ is $(9+5)^{N}$

Now last digit of $(9+5)^{N}$ will be $4^{N}$ because the last digit of (9+5) will only decide the last digit of the exponent

Now, the cyclicity of 4 is 2.

That is last digit of $4^{n}$ is 4 if n is odd and 6 if it is even

Now as N ends with 5

=> N is odd

=> $(9+N)^{N}$ ends with 4 as unit digit.

Note: a%b means the remainder we get when we divide a by b