First show the statement holds for \(n=1\). The left hand side is just 8, and the right hand side is \(4(1)(1+1)=8\), so it's true.
Assume the statement holds for \(n=k\), i.e.
\(8+16+\cdots+8(k-1)+8k=4k(k+1)\)
and use this to show it holds for \(n=k+1\), i.e.
\(8+16+\cdots+8(k-1)+8k+8(k+1)=4(k+1)(k+2)\)
By the assumption above, you have
\(\underbrace{8+16+\cdots+8(k-1)+8k}_{n=k}+8(k+1)=4k(k+1)+8(k+1)=4(k+1)(k+2)\)
so the statement is true for all \(n\ge1\).
