Prove that $1^3 + 2^3 + \cdots + n^3 < n^4$. I am trying to prove the following: $1^3 + 2^3 + \cdots + n^3 < n^4$ if $n \in \mathbb{N}, n>1$ by induction. From there, I am to prove that the sum is $< \frac{n^4}{2}$ if $n>2$. My attempt to do so is as follows:
First, we use $n = 2$ as a basis step since $n \in \mathbb{N}, n>1$ which shows that $1^3+2^3 < 2^4 \Rightarrow 9<16$ which is true. Now, for $n+1$, we have $n^3 + (n+1)^3 < n^4 + (n+1)^4$ by the induction hypothesis. By expanding the two binomials, we have
$2n^3+3n^2+3n+1< 2n^4+4n^3+6n^2+4n+1$. By simplifying our expression, we are left with
$0<n(2n^3+2n^2+3n+1)$ which would always be true with $n \in \mathbb{N}, n>1$. 
I am not sure if this proof would be valid, or if it is logically correct. As for the second part of the question, I would not know where to begin. Could somebody please provide a hint or point out where this goes awry? Any suggestions and/or criticism is welcome. 
I am using the textbook Introduction to Analysis by Arthur Mattuck. 

As the others have said, you went wrong in the induction step. Here's one way to prove what we want without having to expand the binomials:
\begin{align*}
1^3 + 2^3 + \ldots + (n+1)^3 &= \left[1^3 + 2^3 + \ldots + n^3 \right] + (n+1)^3 \\
&< n^4 + (n+1)^3 \qquad\text{by the induction hypothesis} \\
&= n(n)^3 + 1(n+1)^3 \\
&< \color{red}{n}\color{blue}{(n+1)^3} \color{red}{+ 1}\color{blue}{(n+1)^3} \\
&= (\color{red}{n+1})\color{blue}{(n+1)^3}  \qquad\text{factor out the common }(n+1)^3 \text{ term} \\
&= (n+1)^4 \\
\end{align*}
as desired.
