Math induction problem with large numbers I am trying to figure out how to prove $17^{200} - 1$ is a multiple of $10$. I am talking simple algebra stuff once everything is set in place.
I have to use mathematical induction.
I figure I need to split $17^{200}$ into something like $(17^{40})^5 - 1$ and have it as $n = 17^{40}$ and $n^5 - 1$.
I just don't know if that's a good way to start.

Since it seems a bit strange to use induction to solve for particular case ($17^{200} - 1$), and it seems from your question that you want to see this solved via an inductive proof, let's use induction to solve a somewhat more general problem, and recover this particular example as a special case.
So, let's make the conjecture that 
$$4 \mid n \implies 10 \mid (17^n - 1).$$
This means we want to show 10 divides $17^{4k} - 1$ for all integers $k$.  Now, the base case is $k = 1$ and we have
$$ 17^{4k} - 1 = 17^4 - 1 = 83520 = 10 * 8352.$$
Indeed, the hypothesis holds in the base case.
Now, assume the statement is true for some integer $m$.  We then have
\begin{align*}
 &&17^{4m} - 1 &= 10 \cdot a &\text{for some } a \in \mathbb{Z} \\
 \implies && 17^4 (17^{4m} - 1) &= 10 \cdot a \\
 \implies && 17^{4m+1} - 83521 &= 10 \cdot a \\
 \implies && 17^{4m+1} - 1 &= 10\cdot a + 83520 \\
 \implies && 17^{4m+1} - 1 &= 10\cdot (a + 8352) 
\end{align*}
Thus, 10 divides $17^{4m+1}$ if it divides $17^{4m}$; hence, we have shown that 10 divides $17^{4k}$ for every integer $k$.
Since 200 is a multiple of 4, the problem at hand is then solved as a special case of this theorem.
