How can you prove $1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$ by induction? Can you provide the steps and corresponding explanations to prove the following predicate by induction?
$$P(n) := 1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$$
I've done some work on it myself by attempting to show that $\frac{k^4}{4} < \frac{(k + 1)^4}{4}$ for the RHS, but I don't understand exactly what I am doing.
Thank you.
Notice: This is not a homework question. I'm attempting to self-study Calculus over the Summer.

Let's try a generalization.
$P(n) := \sum_{k=1}^{n-1} k^{m-1} < \frac{n^{m}}{m} < \sum_{k=1}^{n} k^{m-1}
$
Base case.
For $n=1$
this is
$0 < \frac1{m} < 1
$
which is true.
Induction step.
Suppose true for $n$.
$P(n) := \sum_{k=1}^{n-1} k^{m-1} < \frac{n^{m}}{m} < \sum_{k=1}^{n} k^{m-1}
$
Want to show that
$P(n) \implies P(n+1)$.
First, the left inequality of $P(n+1)$,
which is
$\sum_{k=1}^{n} k^{m-1} < \frac{(n+1)^{m}}{m}
$.
$\begin{array}\\
\sum_{k=1}^{n} k^{m-1}
&=\sum_{k=1}^{n-1} k^{m-1}+n^{m-1}\\
&<\frac{n^m}{m}+n^{m-1}\\
&=\frac{n^{m}+mn^{m-1}}{m}\\
\end{array}
$
so we are done if
$n^m+mn^{m-1}
\lt (n+1)^m
$
or,
dividing by $n^m$,
$1+m/n
\lt (1+1/n)^m
$
and this follows from
Bernoulli's inequality.
Next, the right inequality of $P(n+1)$,
which is
$\sum_{k=1}^{n+1} k^{m-1} > \frac{(n+1)^{m}}{m}
$.
$\begin{array}\\
\sum_{k=1}^{n+1} k^{m-1}
&=\sum_{k=1}^{n} k^{m-1}+(n+1)^{m-1}\\
&>\frac{n^m}{m}+(n+1)^{m-1}\\
\end{array}
$
so we are done if
$\frac{n^m}{m}+(n+1)^{m-1}
\ge \frac{(n+1)^m}{m}
$
or
$n^m
\ge (n+1)^m-m(n+1)^{m-1}
=(n+1)^{m-1}(n+1-m)
$.
or
$1
\ge (1+1/n)^{m-1}(1-(m-1)/n)
$
or
$(1+1/n)^{m-1}
\le \frac1{1-(m-1)/n}
$.
This requires its own proof.
which we will do
by induction on $m$.
For $m=1$
this is
$1
\le 1$
which is true.
Suppose it is true for $m$
where $m < n-1$.
Then
$(1+1/n)^{m}
=(1+1/n)^{m-1}(1+1/n)
\le \frac1{1-(m-1)/n}(1+1/n)
$.
We want 
$\frac1{1-(m-1)/n}(1+1/n)
\le \frac1{1-m/n}
$
or
$(1-m/n)(1+1/n)
\le 1-(m-1)/n
$
or
$1-(m-1)/n
\ge 1-(m-1)/n+m/n^2
$
which is true.
Therefore,
if $m < n-1$,
or $n > m+1$,
the right side is true.
