$1+2+4+\dots+2^{2015} = 2^{2016}-1$ . As $15 = 2^4-1$ , set $y = 2^4$ . We will get that $\frac{2^{2016}-1}{y-1} = \frac{y^{504}-1}{y-1} = 1+y+y^2+\dots+y^{503}$ Hence, $15|1+2+4+\dots+2^{2015}$ , with no remainders.

$1+2+4+\cdots +2^{2015}=2^{2016}-1$

$2^{2016} \equiv (2^{4})^{504} \equiv 1^{504} \equiv 1 \pmod{15}$

Then $2^{2016}-1 \equiv 0 \pmod{15}$

Finally $\boxed{15|(1+2+4+\cdots+2^{2015})}$

1+2+....+2^2015=2^2016-1
Last digit of 2^2016=6(as 2 has cyclicity of 4(2,4,8,6)) so 2^2016-1 ends with 5 as last digit...so it's divisible by 5

next to check divisible by 3 we'll find remainder of 2^2016-1. 2%3=-1 so (2^2016)%3=(-1)^2016=1 and -1%3=-1 so (2^2016-1)%3=1-1=0...it's divisible by 3

number is divisible by both 5 and 3...so it's divisible by 15

The sum of the series in binary is a string of two thousand and sixteen 1's. Fifteen in binary is 1111. Obviously 1111 will divide the sum since four divides 2016 (the quotient in binary will be 504 repetitions of "0001").

$\displaystyle \sum_{k=0}^{n-1} 2^k = 2^n - 1$

And $2^n - 1$ is divisible by $2^m - 1$ if m | n . . . m & n being positive integers.

Which works for m=4 here, where n= 2016 = 504m.

So the given sum, which = $2^{2016} - 1$ , is divisible by $2^4 - 1$ = 15.

Put another way,

$2^{2016} - 1 = (2^{4})^{504} - 1 = 16^{504} - 1$

and that is divisible by 16 - 1 = 15.

$1 + 2 + 4 + ... + 2^{20015} = 2^{2016} - 1$ Applying Euler's theorem or Fermat's little theorem we get: $\begin{cases} 2^4 \equiv 1 \mod(5) \Rightarrow 5 \space | \space 2^{2016} - 1 \\ 2^2 \equiv 1 \mod(3) \Rightarrow 3 \space | \space 2^{2016} - 1 \end{cases}$ . Now, $\gcd(3,5) = 1 \Rightarrow 15 \space | \space 2^{2016} - 1$