Answer
See the proof below.
Work Step by Step
In order to show that the formula holds we must show that it is true for $n=1$ and that if its is true for $n=k$ then, it is true for $n=k+1$.
$\textbf{1.}$ The formula holds for $n=1$ since
$$1=\displaystyle\frac{1^2\left(1+1\right)^2}{4}.$$
$\textbf{2.}$ Suppose the formula holds for $n=k$, that is, suppose that
$$1^3+2^3+\ldots+k^3=\displaystyle\frac{k^2\left(k+1\right)^2}{4}$$
Then, by adding $\left(k+1\right)^3$ to both sides of the previous equation we obtain that
$$1^3+2^3+\ldots+k^3+\left(k+1\right)^3=\displaystyle\frac{k^2\left(k+1\right)^2}{4}+\left(k+1\right)^3$$
Now, observe that
$$\displaystyle\frac{k^2\left(k+1\right)^2}{4}+\left(k+1\right)^3=\left(k+1\right)^2\left(\displaystyle\frac{k^2}{4}+\left(k+1\right)\right)$$
$$\left(k+1\right)^2\left(\displaystyle\frac{k^2+4k+4}{4}\right)=\left(k+1\right)^2\left(\displaystyle\frac{\left(k+2\right)^2}{4}\right)=$$
$$\displaystyle\frac{\left(k+1\right)^2\left(k+2\right)^2}{4}=\displaystyle\frac{\left(k+1\right)^2\left(\left(k+1\right)+1\right)^2}{4}.$$
Thus,
$$1^3+2^3+\ldots+k^3+\left(k+1\right)^3=\displaystyle\frac{\left(k+1\right)^2\left(\left(k+1\right)+1\right)^2}{4}.$$
and therefore, the formula holds for $n=k+1$ whenever it is true for $n=k$.
Hence, the mathematical induction principle guarantees that the formula is true for every positive integer $n$.
