\displaystyle $\begin{align*}
\sum_{k = 1}^n \left( (k + 1)^4 - k^4 \right) & = \sum_{k = 1}^n \left( 4 k^3 + 6 k^2 + 4 k + 1 \right) \\
(n + 1)^4 - 1^4 & = 4 \sum_{k = 1}^n k^3 + 6 \sum_{k = 1}^n k^2 + 4 \sum_{k = 1}^n k + \sum_{k = 1}^n 1 \\
(n + 1)^4 - 1 & = 4 \sum_{k = 1}^n k^3 + 6 \cdot \frac{n (n + 1) (2 n + 1)}{6} + 4 \cdot \frac{n (n + 1)}{2} + n \\
4 \sum_{k = 1}^n k^3 & = (n + 1)^4 - n (n + 1) (2 n + 1) - 2 n (n + 1) - (n + 1) \\
4 \sum_{k = 1}^n k^3 & = (n + 1) \left[ (n + 1)^3 - n (2 n + 1) - 2 n - 1\right] \\
4 \sum_{k = 1}^n k^3 & = (n + 1) \left[ (n + 1)^3 - (2 n + 1)(n + 1) \right] \\
4 \sum_{k = 1}^n k^3 & = (n + 1)^2 \left[ (n + 1)^2 - (2 n + 1) \right] \\
\sum_{k = 1}^n k^3 & = \frac{n^2 (n + 1)^2}{4}
\end{align*}$