 Solution

Substitute $y(x)={\sum }_{j\ge 0}{c}_j{x}^j$ into the Legendre differential equation. Matching the coefficient of $x^j$ givesChoose the parity of $n$: start with nonzero $c_0$ for even $n$, or nonzero $c_1$ for odd $n$, and set the other parity to zero. The numerator is nonzero for every preceding index of that parity below $n$, so $c_n\ne 0$, but at $j=n$ it vanishes. All higher coefficients are zero. This constructs a nonzero polynomial solution of degree exactly $n$.

Its value at $x=1$ cannot be zero. If it had a zero there of multiplicity $k\ge 1$, write $y=(x-1{)}^{k}R$ with $R(1)\ne 0$. The lowest-order term of $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }$ would be $-2k^{2}R(1)(x-1{)}^{k-1}$, whereas $n(n+1)y$ starts at order $k$. The differential equation could not hold. Hence the polynomial can be scaled to satisfy $P_n(1)=1$.

The resulting Legendre polynomials beginEach satisfies both the differential equation and the specified normalization.