Determine what's correct, fix all the mistakes, and assign a letter grade: $A,B,C,D,F.$

Question: Prove by induction that $n6n>6n={77}^3=343$

$77654321=5040343<{50407}^3<7!k>6(k+1{)}^3<(k+1)(k+1)!(k+1)(k+1)k!(k+1{)}^3(k+1{)}^3=k^3+3k^2+3k+1k>6k+1>7k!+3k^2+3k+1<(k+1)k!k>6k!75040k>67775040=35280169<35280k>6k^3+3k^2+3k+1<(k+1)k!(k+1{)}^3<(k+1)!n>6n^3!$ for all integers $n>6\mathrm{.}3k^2+3k+1!$ holds, thus confirming that:

$k^3+3k^2+3k+1<(k+1)k!$

$So,(k+1{)}^3<(k+1)!$

$By$ induction, $we$ have shown that $n^3!$ for all integers $n>6\mathrm{.}3k^2+3k+1$

And

$7775040=35280$

Clearly,

$169<35280$

$So,$ for $k>6,$ the inequality $3k^2+3k+1!$ holds, thus confirming that:

$k^3+3k^2+3k+1<(k+1)k!$

$So,(k+1{)}^3<(k+1)!$

$By$ induction, $we$ have shown that $n^3!$ for all integers $n>6\mathrm{.}3k^2+3k+1$

Given $k>6$ :

$3k^2+3k+1$

And

$7775040=35280$

Clearly,

$169<35280$

$So,$ for $k>6,$ the inequality $3k^2+3k+1!$ holds, thus confirming that:

$k^3+3k^2+3k+1<(k+1)k!$

$So,(k+1{)}^3<(k+1)!$

$By$ induction, $we$ have shown that $n^3!$ for all integers $n>6.k^3+3k^2+3k+1$

$We$ need $to$ show that $k!+3k^2+3k+1<(k+1)k!$ :

Since $k>6,we$ have $k!75040.$ Thus,

$3k^2+3k+1$

Given $k>6$ :

$3k^2+3k+1$

And

$7775040=35280$

Clearly,

$169<35280$

$So,$ for $k>6,$ the inequality $3k^2+3k+1!$ holds, thus confirming that:

$k^3+3k^2+3k+1<(k+1)k!$

$So,(k+1{)}^3<(k+1)!$

$By$ induction, $we$ have shown that $n^3!$ for all integers $n>6.k^3!.$ Therefore,

$k^3+3k^2+3k+1$

$We$ need $to$ show that $k!+3k^2+3k+1<(k+1)k!$ :

Since $k>6,we$ have $k!75040.$ Thus,

$3k^2+3k+1$

Given $k>6$ :

$3k^2+3k+1$

And

$7775040=35280$

Clearly,

$169<35280$

$So,$ for $k>6,$ the inequality $3k^2+3k+1!$ holds, thus confirming that:

$k^3+3k^2+3k+1<(k+1)k!$

$So,(k+1{)}^3<(k+1)!$

$By$ induction, $we$ have shown that $n^3!$ for all integers $n>6.k^3!$ holds. This $is$ our inductive hypothesis.

$We$ need $to$ show that $(k+1{)}^3<(k+1)!.$

Start $by$ expressing $(k+1)!$ :

$(k+1)(k+1)k!$

Next, consider the expression $(k+1{)}^3$ :

$(k+1{)}^3=k^3+3k^2+3k+1$

Since $k>6,we$ know $k+1>7.By$ the inductive hypothesis, $k^3!.$ Therefore,

$k^3+3k^2+3k+1$

$We$ need $to$ show that $k!+3k^2+3k+1<(k+1)k!$ :

Since $k>6,we$ have $k!75040.$ Thus,

$3k^2+3k+1$

Given $k>6$ :

$3k^2+3k+1$

And

$7775040=35280$

Clearly,

$169<35280$

$So,$ for $k>6,$ the inequality $3k^2+3k+1!$ holds, thus confirming that:

$k^3+3k^2+3k+1<(k+1)k!$

$So,(k+1{)}^3<(k+1)!$

$By$ induction, $we$ have shown that $n^3!$ for all integers $n>6.n^3!$ for all integers $n>6,$ follow these steps:

Base Case: $$

First, verify the base case for $n=7$ :

$7^3=343$

$77654321=5040$

Clearly,

$343<5040$

$So,7^3<7!$

Inductive Step:

Assume that for some $k>6,$ the statement $k^3!$ holds. This $is$ our inductive hypothesis.

$We$ need $to$ show that $(k+1{)}^3<(k+1)!.$

Start $by$ expressing $(k+1)!$ :

$(k+1)(k+1)k!$

Next, consider the expression $(k+1{)}^3$ :

$(k+1{)}^3=k^3+3k^2+3k+1$

Since $k>6,we$ know $k+1>7.By$ the inductive hypothesis, $k^3!.$ Therefore,

$k^3+3k^2+3k+1$

$We$ need $to$ show that