algebra precalculus - Prove by induction $sum_{i=1}^ni^3=frac{n^2(n+1)^2}{4}$ for $nge1$

Prove the following statement $S(n)$ for $n\ge1$:

$$\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}$$

To prove the basis, I substitute $1$ for $n$ in $S(n)$:

$$\sum_{i=1}^11^3=1=\frac{1^2(2)^2}{4}$$

Great. For the inductive step, I assume $S(n)$ to be true and prove $S(n+1)$:

$$\sum_{i=1}^{n+1}i^3=\frac{(n+1)^2(n+2)^2}{4}$$

Considering the sum on the left side:

$$\sum_{i=1}^{n+1}i^3=\sum_{i=1}^ni^3+(n+1)^3$$

I make use of $S(n)$ by substituting its right side for $\sum_{i=1}^ni^3$:

$$\sum_{i=1}^{n+1}i^3=\frac{n^2(n+1)^2}{4}+(n+1)^3$$

This is where I get a little lost. I think I expand the equation to be

$$=\frac{(n^4+2n^3+n^2)}{4}+(n+1)^3$$

but I'm not totally confident about that. Can anyone provide some guidance?