Prove by induction that $21^n-21$ is divisible by $5$ for all natural numbers. 
Prove by induction that $21^n-21$ is divisible by $5$ for all natural numbers.

I have tried this question several times but am always having to use that $0$ is divisible by $5$ in step one
So far I have done;
$21^1-21=0$ which is divisible by 5
then assumed true for n=k 
I cannot think of a way to use these steps for n+1
 A: $21^{n+1}-21= 21^n \cdot 21-21=(21^n-21) \cdot 21+21^2-21=(21^n-21) \cdot 21+420.$
Can you proceed ?
A: Base Case ($n = 1$): $21^1 = 21 = 0$, and $0$ is of course divisible by $5$. 
Induction Hypothesis: Assume $21^k - 21$ is divisible by $5$. Hence, there exists $m \in \mathbb{Z}$ such that $21^k - 21 = 5m$. (Hence, $21^k =  21 + 5m$).
Induction Step: We have: 
\begin{align*}
21^{k+1} - 21 & = 21^k \cdot 21 - 21 \\
& = (21 + 5m) 21 - 21 \\
& = 441 + 5(21m) - 21 \\
& = 420 + 5(21m) \\
& = 5(84) + 5(21m) \\
& = 5(84 + 21m).
\end{align*}
A: You can see that $21^{n+1}-21= 21^{n+1}-21^n+21^n-21=21^n(21-1)+(21^n-21)=20\cdot 21^n + (21^n-21)$
Since $21^{n+1}-21$ can be written as Sum of two multiple of 5 you have the inductive step!
A: Another way of looking at it, as mentioned in the comments, is that as $21$ ends in a $1$ multiplying it by itself will also result in a number which ends in $1$:
\begin{align}21^2&=441\\
21^3&=9261\\
21^4&=194481\\
&\vdots\end{align}
We then subtract $21$ from this, so the resulting number must end in a $0$ and thus is divisible by $5$, as all multiples of $5$ end in a $0$ or a $5$ 
A: Step: $n+1$
$21^{n+1}-21=21(21^n-1)=$
$21((21^n-21)+(21-1))=$
$21(21^n-21)+21\cdot 20$;
Since $(21^n-21)$ is divisible by $5$ (hypothesis), so is the first summand, and so is the second summand $21\cdot 20.$
