Question 670006
To solve a question like this, one looks at {{{patterns}}} of powers and a {{{cycle}}} of {{{length}}}. 

{{{7^1 =7}}}.................unit's dig. {{{7}}}
{{{7^2 = 49}}}...............unit's dig. {{{9}}}
{{{7^3 = 343}}}..............unit's dig. {{{3}}}
{{{7^4 = 2401}}}.............unit's dig. {{{1}}}...pattern:{{{7}}},{{{9}}},{{{3}}},{{{1}}}
{{{7^5 = 16807}}}............
{{{7^6 = 117649}}}...........
{{{7^7 = 823543}}}...........
{{{7^8 = 5764801}}}...........pattern is repeating
{{{7^9 = 40353607}}}......
{{{7^10 =282475249}}}.....
{{{7^11 =1977326743}}}.....
{{{7^12 =13841287201}}}......pattern is repeating

So the unit's digits of powers of {{{7}}} follow a cycle of length {{{4}}}: {{{7}}},{{{9}}},{{{3}}},{{{1}}} ... ... 

so let's find the remainder when we divide{{{157}}} by {{{4}}}: {{{157 = 39*4 + 1 }}}
Since the remainder is {{{1}}}, and {{{7^157}}} has the same unit's digit as {{{7^1}}}, which is {{{7}}}.