proof by induction: $7^{2n} - 2^n$ is always divisible by $47$ as per the title, i need to prove by induction that the expression "$7^{2n} - 2^n$" is always divisible by $47$.  

base case is fine: $7^{2*1} - 2^1 = 47$. 

Induction step gets lost quickly, any ideas?

step 1: $7^{2(n+1)} - 2^{(n+1)}$ 

step 2:  $7^2 * 7^{2n} - 2^1 * 2^{n}$ 

step 3: ???

step 4: profit.
Step 1 is not “$7^{2(n+1)} - 2^{n+1}$.”  You're thinking in terms of expressions and you want to think in terms of sentences.  This is one half of the conclusion to an if-then statement you want to prove:


  For any $n$, if $7^{2n} - 2^n$ is divisible by $47$, then $7^{2(n+1)} - 2^{n+1}$ is also divisible by $47$.


Step 0 is the base case, which you have shown.  

Step 1 is “Suppose there exists a positive integer $n$ such that $7^{2n} = 2^n$.”  

Step 2 is: you need to show that $7^{2(n+1)} - 2^{n+1}$ is divisible by $47$.

You've got a few answers on how to show that.  Essentially it's the old add-and-subtract-the-same-thing trick.
\begin{align*}
    7^{2n+2} - 2^{n+1} &amp;= 7^2\cdot 7^{2n} - 2\cdot 2^{n} 
\\&amp;= 7^2 \cdot 7^{2n} - 7^2 \cdot 2^n + 7^2 \cdot 2^n - 2\cdot 2^n
\\&amp;= 49 (7^{2n} - 2^{n}) + 47 \cdot 2^n
\end{align*}
If $7^{2n} - 2^{n} = 47k$ for some integer $k$, then
$$
    49 (7^{2n} - 2^{n}) + 47 \cdot 2^n = 49\cdot 47k + 47 \cdot 2^n = 47(49k + 2^n)
$$
in other words, also a multiple of $47$.

Step 3 is “Therefore, by induction, $7^{2n} - 2^n$ is divisible by $47$ for all positive integers $n$.”

If you follow these steps, and apply them to all proofs by induction, you will indeed profit.
Step 1:  $7^{2(n+1)} - 2^{n+1}$

Step 2: $7^2*7^{2n} - 2*2^n$.

Step 3: $(47 + 2)7^{2n} - 2*2^n=47*7^{2n} + 2(7^{2n} - 2^n)$

Step 4: Profit.

(BTW... you miswrote this as $7^{2n} - 2^{2n}$ which is not true.)