Is this application of induction correct? Prove that the sum of first $n$ natural numbers is $\frac{n(n+1)}{2}$
So I verify the base case for $n=1$.
Now I assume that the proposition is true for all natural numbers up to $n=k$. So my question is, is this assumption valid?
 A: You can assume anything you want in a proof; no sense in questioning that assumption.  But if that assumption leads to a certain result, you can infer something from the very fact that the assumptions leads to some other result. 
So it is with induction: it is the fact that you can show that the proposition is true for  $k+1$ once you make the inductive hypothesis that the proposition is true for all numbers up to $k$ that, together with the inductive base, allow you to prove that the proposition holds for all $n$. But there is no sense in questioning the validity of the inductive hypothesis. 
And, if you're still worried about the truth of the inductive hypothesis, do note that once the induction is completed, you will have shown the proposition to be true for all $n$, and thus show that the inductive hypothesis was right all along. ... which always makes induction feel like it is a circular argument, but it isn't.
A: You don't need strong induction here. Assume the sum of the first $k$ natural numbers is $\dfrac{k(k+1)}2$. Then the sum of the first $k+1$ natural numbers is $(1+2+\cdot\cdot\cdot+k)+k+1=\dfrac{k(k+1)}2+k+1=\dfrac{(k+1)(k+1+1)}2$, which is what we wanted to show.
