Induction problem clarification

Question

here's the problem I'm doing:

Prove that for all integers $n$ with $n \geq 1$, we have $n \cdot 6^n \leq (n+10)!$

I don't understand how to get from [$6 \cdot (k + 10)! + 6^{k+1}$] to $k \cdot (k + 10)! + 11 \cdot (k + 10)! $.

Base Case:

Let $ n = 1$.

Then, $LHS = 1 \cdot 6 = 6$

$RHS = (1 + 10)!$

Clearly, $6 \leq 10!$ and hence, the inequality is satisfied for the base case.

Inductive Hypothesis:

Let us assume that for $n = k$, we have $k \cdot 6k \leq (k + 10)!$

Inductive Step:

Now, we would need to prove that for $n = k + 1$, the inequality holds true.

Proof:

$= (k + 1) \cdot 6k+1= 6k * 6^{k} + 6^{k+1} \leq 6*(k + 10)! + 6^{k+1} \leq k * (k + 10)! + 11 * (k + 10)! \leq (k + 11)(k + 10)! \leq (k + 11)!$

Answer 1

I was confused by the OP's work, so I didn't focus that closely on it. Anyway: as $n \to (n+1),$
the LHS increases by a factor of $6\frac{n+1}{n}$
while the RHS increases by a factor of $(n+1).$

For $n \geq 7, \;6\frac{n+1}{n} < 6\frac{n+1}{6} = (n+1).$
Therefore, for $n \geq 7,$ the LHS is increasing by a smaller factor than the RHS.

Thus, you simply have to manually check that the assertion is true for $n \,\in \,\{1,2,3,\cdots,7\}.$

Then, use can use $n=7$ as the base case an apply induction against all $n > 7.$

$\underline{\text{Addendum}}$
IamWill's answer is better than mine, because he found analysis that allows the induction to begin at the base case of $n=1,$ rather than $n=7.$

Answer 2

You can do it more straightfoward. Because you are assuming $k \ge 1$, you have obviously $6^{k} \le k6^{k} \le (k+10)!$. Thus: $$6(k+10)!+6^{k+1} = 6[(k+10)!+6^{k}] \le 6[(k+10)!+(k+10)!] = 12(k+10)! \le (k+11)(k+10)!$$