= Solution
The original PDF contains this subpart; the local TeX has omitted it. Write $a_n$ for the <leading coefficient> of $P_n$. The <Legendre polynomial recurrence relation> and the starting <polynomials> give $a_0=a_1=1$, and induction gives
$$
a_{n+1}=\frac{2n+1}{n+1}a_n>0.
$$
Indeed, $xP_n$ has <degree of a polynomial> $n+1$, while $P_{n-1}$ has smaller <degree of a polynomial> and cannot cancel its leading term. Hence
$$
\boxed{\deg P_n=n}.
$$
At $x=1$, the same <linear recurrence relation> gives $(n+1)P_{n+1}(1)=(2n+1)P_n(1)-nP_{n-1}(1)$. Starting from $P_0(1)=P_1(1)=1$, induction yields
$$
\boxed{P_n(1)=1\quad(n\geq0)}.
$$
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