Proving $\sum_{k=1}^nk^3 = \left(\sum_{k=1}^n k\right)^2$ using complete induction I tried to prove the following statement using complete induction but I couldn't manage to solve it because I got a complex notation eventually.
The statement is the following:
$$\sum_{k=1}^nk^3 = \left(\sum_{k=1}^n k\right)^2$$
I hope someone can help me out!
 A: Hint:prove following statements  by induction $$\sum_{k=1}^nk^3 =\left(\frac{n(n+1)}{2}\right)^2$$
 $$\sum_{k=1}^n k=\frac{n(n+1)}{2}\to\left(\sum_{k=1}^n k\right)^2=\left(\frac{n(n+1)}{2}\right)^2$$ then we conclude
$\sum_{k=1}^nk^3 = \left(\sum_{k=1}^n k\right)^2$
A: Let's replace $n$ by $n+1$ on both sides and see if it works out:
$$\sum_{k=1}^nk^3 + (n+1)^3= (1+2+\ldots+n+(n+1))^2=\\
\left(\sum_{k=1}^n k\right)^2+2(1+2+\ldots+n)(n+1)+(n+1)^2$$
Since $(1+2+\ldots+n)=\frac{n(n+1)}{2}$ we get
$$(n+1)^3=2\frac{n(n+1)}{2}(n+1)+(n+1)^2=(n+1)^3\checkmark$$
which proves the equality.
A: The base case is easy. To perform the induction step, note that $$\begin{align}\left(\sum_{k=1}^{n+1}k\right)^2 &= \left(n+1+\sum_{k=1}^nk\right)^2\\ &= (n+1)^2+2(n+1) \sum_{k=1}^nk+ \left(\sum_{k=1}^nk\right)^2\\ &= (n+1)^2+ 2(n+1) \sum_{k=1}^nk+\sum_{k=1}^nk^3\end{align}$$ by induction hypothesis. It remains only to show that $$ (n+1)^2+ 2(n+1) \sum_{k=1}^nk=(n+1)^3.\tag{$\star$}$$ Equivalently, we must show (or use the fact, if you know it) that $$\sum_{k=1}^nk=\frac{n(n+1)}2.\tag{$\heartsuit$}$$ Do you see why $(\star)$ and $(\heartsuit)$ are equivalent?
