What I've done so far:

For n = 1

\(\displaystyle 7^1-2^1=7-2=5\) (true that it is divisible by 5)

For n = k

\(\displaystyle 7^k-2^k=5a\) (assumed to be true that it is divisible by 5)

For n = k + 1

\(\displaystyle 7^{k+1}-2^{k+1}=7^k\cdot7-2^k\cdot2=7(7^k-2^k)+12\cdot2^k=7(5a)+12\cdot2^k\)

This is where the problem lies. How can I show that \(\displaystyle 12\cdot2^k\) is divisible by 5?