Not following what's happening with the exponents in this proof by mathematical induction. I'm not understanding what's happening in this proof. I understand induction, but not why $2^{k+1}=2*2^{k}$, and how that then equals $k^{2}+k^{2}$. Actually, I really don't follow any of the induction step - what's going on with that algebra? Thanks!
Let $P(n)$ be the statement
$2^n > n^2$, if $n$ is an integer greater than 4.
Basis step:
$~~~~~P(5)$ is true because $2^5 = 32 > 5^2 = 25$
Inductive step:
$~~~~~~$Assume that $P(k)$ is true, i.e. $2^k > k^2$. 
We have to prove that $P(k+1)$ is true. Now
$$\begin{aligned}
2^{k+1} & = 2 \cdot 2^k\\
& > 2 \cdot k^2\\
& = k^2 + k^2\\
& > k^2 + 4k\\
& \geq k^2 +2k+1\\
& = (k+1)^2
\end{aligned}$$
$~~~~~$ because $k>4$
$~~~~~$ Therfore $P(k+1)$ is true
Hence from the principle of mathematical induction the given statement is true. 

I have copied the induction step table format from your question and will reformat it as I had to do in my high school geometry class. I find it very helpful to get away from the table as you gave it.
Set $P(n)$ equal to the statement that 2^n>n^2 for all n>4. Note the basis step 2^4>4^2 is true.
$~~~~~~$Assume that $P(k)$ is true, i.e. $2^k > k^2$. 
We have to prove that $P(k+1)$ is true. Now
\begin{array}[10,2]{}
STATEMENT & REASON \\
2^{k+1}  = 2 \cdot 2^k & \text{Induction step  of  the  definition  of }2^n \\
2^k > k^2  & \text{This is the induction hypothesis.} \\ 
2 \cdot 2^k > 2 \cdot k^2= k^2 + k^2 & \text{Multiplying by a positive number preserves inequalities} \\ & \text{ and the distributive law.} \\
k^2 + k^2 > k^2 + 4k & k>4\text{ by hypothesis, so } k^2 >4k \text{ and then } k^2 + 4k > k^2 + 2k +2\\ & \text{and the standard rules of inequalities used above apply.} \\  
k^2 + 4k \geq k^2 +2k+1 & 4k > 2k+1 \text{ since  }2k-1>0 \\
k^2 +4k +1 =(k+1)^2 & \text{Multiplication distributes over addition.} \\
\end{array}
$~~~~~~$
This shows $P(k)\implies P(k+1)$, and the principle of finite induction completes the proof that $P(n)$ is true for all $n>4$ .
