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. 
 A: You go astray at the beginning of the induction step. Your induction hypothesis is that 
$$1^3+2^3+\ldots+n^3<n^4\;,\tag{1}$$
and you want to show that 
$$1^3+2^3+\ldots+n^3+(n+1)^3<(n+1)^4\;.$$
In other words, you don’t have $n^3+(n+1)^3<n^4+(n+1)^4$ by the induction hypothesis: what you have is that
$$\Big(1^3+2^3+\ldots+n^3\Big)+(n+1)^3<n^4+(n+1)^3\;,$$
obtained by adding $(n+1)^3$ to both sides of $(1)$. To complete the induction step, you need to show that $n^4+(n+1)^3\le(n+1)^4$, and what you wrote, though incorrect, shows that you probably can do that just fine.
A: 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.
A: You don't quite have the induction hypothesis correct. The hypothesis is that
$$\sum\limits_{k = 1}^n k^3 < n^4$$
Now the goal is to show that, assuming the hypothesis,
$$\sum\limits_{k = 1}^{n + 1} k^3 < (n + 1)^4$$
Now we consider the left hand side, getting:
$$\sum\limits_{k = 1}^{n + 1} k^3 = \sum\limits_{k = 1}^{n} k^3 + (n + 1)^3 < n^4 + (n + 1)^3$$
where the inequality uses the hypothesis. So really, your goal reduces to showing that
$$n^4 + (n + 1)^3 < (n + 1)^4$$
A: Yet another solution. 
Assuming $n$ is even, (by making a little bit change, it will fit odd too)
$$ 1^3 + 2^3 + \dots + (n-1)^3 + n^3 $$ 
$$< $$
$$ (n/2)^3 + (n/2)^3 + \dots + n^3 + n^3 $$
$$ < $$
$$ (n/2)*(n/2)^3 + (n/2)*n^3 = n^4/16 + n^4/2 = \frac{9}{16}n^4$$
$$ < $$
$$ n^4$$
