Mathematical induction on a finite subset of $\mathbb{N}$ I'm trying to prove the following using induction (or well-ordering principle). (Induction says that whenever $ 0 \in S $ and $ n \in S \implies (n+1) \in S$, then $S = \mathbb{N} $. )
Let $ \{ P_{k} \}$  be a sequence of propositions such that
$P_{1}$ is true
The truth of $P_{i-1} $ implies the truth of $P_{i}$ for all $(i-1) \in \{ 1,2,...(n-1) \} $  
Then show that ${P_{k}}$ is true for all $ k \in \{ 1,2,...n \} $  
Clearly, this is a modified form of induction (perhaps a weaker form) restricted to the first $n$ naturals, so the knowledge of induction should easily allow me to prove this. Yet I am stuck, what's the trick? 
 A: We use the Well-Ordering Principle.
Let $T = \{k \in \{1, 2, 3, \ldots, n\} \mid P_k~\text{is false}\}$.  We will show that $T = \emptyset$.  
Suppose otherwise.  Then since $T \subseteq \mathbb{N}$ and $T \neq \emptyset$, there exists a least element $m \in T$ such that $P_m$ is false.  Since $P_1$ holds, $m \neq 1$.  Since $m > 1$ and $m - 1 < m \leq n$,  $P_{m - 1}$ holds.  However, $P_{m - 1} \implies P_m$ holds.  Thus, $m \notin T$, contrary to our assumption that $m$ is the least element of $T$.  Hence, $T$ does not have a least element, so $T = \emptyset$.  Therefore, if $P_1$ holds and $P_{k - 1} \implies P_k$ for all $k \in \{1, 2, 3, \ldots, n - 1\}$, $P_k$ holds for each $k \in \{1, 2, 3, \ldots, n\}$.     
A: On easy way to use the induction principle is to create a new proposition which hold for all natural numbers.
Consider $Q_k$ to be the proposition $P_k$ if $k\leq n$, and other wise i.e. if $k>n$ then let $Q_k$ be the statement $k>n$.
Now you may prove that $Q_k$ hold using induction. Notice that you will need to divide into cases in the induction step depending on if the number you do incution over is greater than $n $ or not.
A: Now we have the truth of $P_{i-1}$ implies the truth of $P_i$ for all $(i-1) \in \{ 1,2,...,n \}$, and we repeat.
