Proof by Induction: $∀n ≥ 5, 2^n + 2n < n!$ I'm having trouble understanding how to solve this question. 
Proof by induction:  $∀n ≥ 5, 2^n + 2n < n!$
I don't understand how they got those steps that I highlighted in the picture attached below.
Any help is appreciated. 

Thanks!
 A: You are assuming $2^k + 2k < k!$ and tring to prove $2^{k+1} + 2(k+1) < (k+1)!$. Because you are assuming $2^k + 2k < k!$ then 
$$2^k + 2k < k! \Rightarrow 2(2^k + 2k) < 2(k!) \Rightarrow 2(2^k + 2k) +2 < 2(k!) + 2$$
and this is the first highlighted step. 
For the second one, because $k\geq 5$ (so $k>2$): 
$$2(k!)+2 < 2(k!) + k! < 3(k!) < (k+1)(k!)$$
A: So first, we need to understand what does one actually do, in induction we first assume that the hypothesis is true for some iteration $k$ and then show that it also holds for $k+1$. After we have checked for some particular value and followed the above step, induction says that the statement is true for any iteration. 
We have:
$\forall\ n\ge 5$, W.T.S. $2^n+2n < n!$.
First, see that for $n= 5, 2^5 + 2\cdot 5 = 32+10 = 42 < 5! = 120$. So the statement is true for some value, namely $5$.
Next, we assume it to be true for $n = k$, and try to show for $n =k+1$.
$2^{k+1}+ 2(k+1) = 2\cdot 2^k + 2k +2  < 2(2^k+2k)+2 < 2k! + 2$ (because we assumed that statement is true for $k$) $< 2k! + k!$ (since $k\ge 5$) <3k! < (k+1)! 
Hence we prove the statement for $n =k+1$ and hence it is true for all $n$ by induction. 
Hope this helps.
