Prove: P(n): (1^3)+(2^3)+...+(n^3)=(1+2+...+n)^2 for all natural numbers n

Attempt so far:

1. Basis Step - P(1) is true since 1^3=1^2

2. Induction Step - NTS P(k)--> P(k+1)

3. Suppose (1^3)+(2^3)+...+(k^3)=(1+2+...+k)^2 is true for some k, then we NTS P(k+1) is true.

4. P(k+1): (1^3)+(2^3)+...+(k^3)+((k+1)^3)=(1+2+...+(k+1))^2

5. From our supposition, the LHS becomes:

6. P(k+1): (1+2+...+k)^2 + (k+1)^3 = (1+2+...+(k+1))^2

This is where I'm stuck. Also, on step four, why does the LHS have two k's (the k and k+1), while the RHS only has one? There's a similar example that does this, but I don't know why I did it here.

Any help would be appreciated. Thanks!