Proving orthogonality of Legendre polynomials

Question

I know that such question has already been answered but what I am trying to figure out is that whether theres a more elementary way of doing so. I know the proof where we take the help of Legendre differential equation ie $$(1-x^2)y''-2xy'+n(n+1)y=0.$$ But I do not find it intuitive as that's not the proof that will strike you when you have just started dealing with these polynomials.

My Attempt

So my first approach towards this problem of proving $$\int _{-1}^{1} P_m (x) P_n (x) dx=0$$ where $P_i(x)$ is the $i$th degree Legendre polynomial is as follows:

Starting with $$P_m (x) = a \frac{d^m}{dx^m}(x^2-1)^m=a (x-1)^m(x+1)^m,$$ doing similar thing for $n$ and then somehow using Leibniz rule and saying that at $x=-1,1$ since $(x+1),(x-1)$ are $0$ respectively that's why the whole integral sums up to zero.

Answer 1

So preceding with your idea of using Rodrigue's formula for the Legendre polynomials $P_n (x)$, namely $$P_n (x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} [(x^2 - 1)^n],$$ I will start by considering the integral $$\int_{-1}^1 f(x) P_n (x) \, dx.$$

Substituting Rodrigues formula for $P_n (x)$ into the above integral followed by integrating by parts $n$ times, recoginising the term $(x^2 - 1)$ repeatedly cancels at either of the end-points, one will be left with $$\int_{-1}^1 f(x) P_n (x) \, dx = \frac{(-1)^n}{2^n n!} \int_{-1}^1 f^{(n)}(x) (x^2 - 1)^n \, dx.\tag1$$

If $f(x) = P_m (x)$ where $m$ is an integer such that $0 \leqslant m < n$, since $P_m (x)$ is a polynomial of degree $m$ which is less than $n$, we have $$f^{(n)}(x) = \frac{d^n}{dx^n} [P_m(x)] = 0,$$ and from (1) we see that $$\int_{-1}^1 P_m (x) P_n (x) \, dx = 0, \quad m \neq n,$$ as required.