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. 

The arithmetic has already been explained, but not the intuition motivating the induction. We desire to prove by induction that the function $\rm\ \color{#C00}{2^n - n^2 =\, f(n)}\,$ is positive for $\rm\:n \ge 5.\:$ Notice
$$\begin{eqnarray}\rm f(n\!+\!1) &=&\rm\ 2\cdot \color{#C00}{2^n} - (n\!+\!1)^2 =\, 2\,(\color{#C00}{f(n)+n^2})-(n\!+\!1)^2\\
\rm \Rightarrow\ \ f(n\!+\!1) &=&\rm\ 2\: f(n) + (n\!-\!1)^2\! - 2\end{eqnarray}$$ 
Now the induction step is clear: $\rm\ f(n)> 0\,\Rightarrow\, f(n\!+\!1) > 0\,$ since $\rm\,f(n\!+\!1)\,$ is a sum of terms $\,> 0,$ namely, the first rhs summand $\rm\,2\,f(n)>0\:$ since $\rm\:f(n)>0\:$ by the induction hypothesis; $ $ further, the second rhs summand $\rm\:(n\!-\!1)^2\!-2 > 0\:$ for $\rm\:n \ge 3\ $ (by $\rm\,x^2\,$ is increasing, or by induction).    
More generally one can easily prove by induction that if a function on $\,\Bbb N\,$ satisfies a recurrence of the form $\rm\: f(n\!+\!1) =\, a(n)\, f(n) + b(n)\:$ and $\rm\,f(n_0)> 0\:$ then $\rm\,f\,$ remains positive for all $\rm\:n \ge n_0,\:$ assuming that both $\rm\:a(n)>0\,$ and $\rm\,b(n) > 0$ for $\rm\:n > n_0.\:$ It's really nothing more than the obvious induction that an increasing function stays larger than its initial value, which amount essentially to iterating the transitivity law for inequalities $\rm\:f_{n+1} > f_n > \cdots > f_{0}\:\Rightarrow\: f_{n+1} > f_{0}.$
Many induction proofs can be cast in this form. For example, products and exponentials of polynomials satisfy such recurrences. The method often succeeds because it replaces hairy exponential arithmetic by much simpler polynomial arithmetic (verifying positivity of the polynomials $\rm\:a(n),b(n),\:$ which requires no insight since there is a simple algorithm).
This is a simple generalization of telescopy - about which much has been written here. 
