Showing that $3n
How can we show that:
  $$3n< n!$$
  whenever $n$ is an integer such that $n \geq 7$ ?  
I was thinking that we can prove this by showing that such case is true with any integer above 7, but then I remembered that I think that only works when we are disproving something.  Is there any way to show this another way?  Thanks!
Edit: I was thinking along the lines of using base case of 7 and plugging in examples to proof?
 A: If you are looking for a less efficient way to approach this proof, try proof by induction, if for no other reason than practice with inductive proofs:
Let $P(n)$ be the assertion: $$3n<n!,\;\;n\geq 7\tag{P(n)}$$


*

*Establish the assertion is true for the base case $n = 7$, by computing $$3\cdot 7 < 7! \tag{P(7)}$$ and confirming the inequality holds for $n = 7$.

*Assert the inductive hypothesis: assume the assertion $P(n)$ is true for $n = k$: that is, assume $$3k < k!\tag{inductive hypothesis: $P(k)$}$$

*Then make the inductive step by showing that, using the inductive hypothesis, it follows that $$3(k+1) < (k + 1)!\tag{P(k+1)}$$
If you can prove $P(k + 1)$ follows from $P(k)$, then together with the truth of the base case, you conclude: therefore $P(n)$ is true.
The task is noting that $$3(k+1)=\color{blue}{\bf 3k}+3<\color{blue}{\bf k!}+3<k!(k+1) = (k + 1)!$$ (note we used the inductive hypothesis $\color{blue}{\bf 3k<k!}$). Hence $P(k+1)$ is indeed true, given $P(k)$.
With the base case, you are done, you have proven that $P(n)$ is true for $n\geq 7$.
A: Hint: $(n-1)!>3,\forall n\ge4$
Multiplying with n on both sides we get $n!>3n,\forall n>3$.
A: Use induction to prove this
Take your hypothesis as $P(n) : 3n < n!$ 
First show that $P(7)$ is true
Next assume $P(n)$ is true and show that $P(n+1)$ is true.
If you are still confused why this is a valid proof I can explain, just comment and let me know
$$
P(n+1) : 3(n+1) < (n+1)! \\
3n + 3 < n!(n+1)
$$
By the inductive hyposthesis we know
$$
3n < n!
$$
So plug and chug and look at what your inequality gets you...
(This is exactly what @amWhy has, I just tried to be more precise I guess... anyways she's completely right I'm just trying to clear up your question on her answer)
