Find the multiplicative order of $15$ mod $ 257$ could someone please help me with this problem?  Thanks in advance! 

Find the multiplicative order of $15$ modulo $257$.
$$15^2\equiv-2^5\pmod{257}$$

Now $2^8\equiv-1\pmod{257}$

$\implies$ord$_{257}2=16$

$\implies$ord$_{257}2^5=\dfrac{16}{(16,5)}=16$

$\implies$ order of $-2^5$ modulo $257$ will be $16$ as for $(-2^5)^{2m}=(2^5)^{2m}$ for any integer $m$

$\implies$ord$_{257}15=2\cdot16$
Elementary theory in modular arithmetic says that $x^p-x\equiv0\mod p$ for any prime $p$. In this case, we have $15^{257}-15\equiv0\mod257$ so $15^{256}\equiv1\mod257$. The order of $15$ then, is a divisor of $256=2^8$. So it must be one of the numbers $1,2,4,8,16,32,64,128,256$.