1.
1^3 + 2^3 + ... + n3 = (n(n+1)/2 )^2 for any positive integer n.
2.
3^n < n! for all n > 6 with n ∈ N.
mathematical induction to prove these?
Yes that's the way to go. Without going through all the formalities ...
#1: Assume it holds for $\displaystyle k$. We want to show that it holds for $\displaystyle k+1$ as well, that is, we want to show that $\displaystyle 1^3 + 2^3 + \cdots + k^3 + (k+1)^3 = \left(\frac{(k+1)(k+2)}{2}\right)^2$
So for your inductive step: $\displaystyle {\color{red}1^3 + 2^3 + \cdots + k^3} + (k+1)^3 = {\color{red}\left(\frac{k(k+1)}{2}\right)^2} + (k+1)^3 = \cdots$
#2: Again, assume it holds for $\displaystyle k$ for all $\displaystyle k > 6$, so we're assuming that $\displaystyle 3^k < k!$. We then want to show that it is true for $\displaystyle k+1$ as well, that is, $\displaystyle 3^{k+1} < (k+1)!$
Note that: $\displaystyle 3^{k+1} = 3^k{\color{blue}3^1} < 3^k{\color{blue}(k+1)} < \cdots$
We can assume the blue because we're given that the statement is true for values strictly greater than 6. So obviously, $\displaystyle 3 < 6 < k+1$. Hopefully you'll see the conclusion directly after using the inductive step.