$4^{250}\; mod\; 14$ is equivalent to $ 16^{125}\; mod\; 14$
And $16^{125} = (14 + 2)^{125}$
Every term in expansion of $(14+2)^{125}$ will contain $14$ as a factor other than last term $2^{125}$
So, Problem reduced to $2^{125} \;mod\;14$
Again, it can be rewritten as $2*16^{31} \;mod\;14$ which is equivalent to finding $2^{32} \;mod\;14$
$2^{32}\;mod\;14 = 16^8\;mod\;14 = (14+2)^8\;mod\;14$( Every term except $2^8$ contains a factor $14$.)
Problem reduces to $\color{maroon}{2^8\;mod\;14 = 256\;mod\;14 = 4}$