- There are many examples where this happens with properties concerning integers, where a property for a given integer may not be easy to prove, but the proof for the $n=0$ or $1$ case is easy, and induction is easy as well.

Alain Connes gave a great example of this on the radio you could even explain to a non-mathematician (actually that was the point): if I give you a chocolate bar with $8\times 5$ squares, and you're allowed to cut it by breaking any piece into two pieces along a straight line (not intersecting the squares), what is the way of doing it that requires the least amount of cutting ? For $8\times 5$ it's not clear, but if you generalize to $n\times m$ and use induction, then it becomes quite clear.

Perhaps a more mathematical example of this phenomenon would be quite appreciated: many divisbility questions fit into this, e.g. prove that $2^{2n}+2$ is divisible by $3$

- A different family of examples comes from applications of group theory or ring theory to number theory; for instance Euler's theorem : $a^{\varphi(n)} = 1 \mod n$ if $a\land n = 1$. I have never seen Euler's original proof of this, but seeing how complicated non-group-theoretic proofs of Fermat's little theorem are, I can only imagine that proving this only by number theory must be hard; while Lagrange's theorem is really easy and deals with this immediately if you know Bezout's theorem.

Another example in this family could be the fact that $n\mid \varphi(a^n-1)$; or $n! \mid \displaystyle\prod_{i=0}^{n-1}(q^n-q^i)$ whenever $q$ is a prime power (though for these, unlike with the first family of examples, it may not be clear *how* to go from the specific instance to the generalization)