Power of 2017

Suppose it is false. Then each digit ( $0, 1, 2, 3, \dots, 9$ ) appears at most two times. Since there are $10$ digits and $p^k$ has $20$ digits, each digit appears exactly two times. Hence the digits of the number are $0,0,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9$ in some order. Note that a power of prime number is only divisible by powers of the same prime number. But $p>3$ , so $3\not\mid p^k$ would have to be true, but it is false, because $3\mid 0+0+1+1+2+2+3+3+4+4+5+5+6+6+7+7+8+8+9+9=90$

So it is $\boxed{\text{true}}$ .