The degree $1$ case can be proved by using the fact that a sequence of prime numbers must grow faster than linearly : assume the property isn’t true for a polynomial $P = aX + b$ and let $(u_n)$ be the sequence of its prime values in increasing order. There’s a fixed $n_0$ such that every $n_0$ consecutive values of $P$ contains a prime so $|u_{n+1}-u_n| \le |a| n_0$ and $u_n = O(n)$. But this is impossible since this is a subsequence of the sequence of prime numbers, which grows as $n \ln(n)$.
Now for the general case : take your polynomial $P$ and your number $n$. We aim to find a suitable $a$. Let $p_1< p_2 < \ldots <p_n$ be distinct primes such the $P$ has a root modulo $p_k$ for all $1\le k \le n$ (we can take as many as needed because there are infinitely many such primes. See here for a proof Primes dividing a polynomial), and for each $p_k$, we chose $0 \le r_k < p_k$ such a root. Now we pick $a$ such that for all $k$, we have
$$ a+k \equiv r_k \mod p_k$$
Which is possible thanks to the Chinese remainder theorem. Theses congruences insure that $p_k$ divides $P(a+k)$. Now since we can take $a$ as big as we want (if $a_0$ satisfies the congruences, so does $a_0+N \times (p_1 p_2 \ldots p_n)$ for any $N$), we can insure $|P(a+k)| > p_k$, which implies that it must be composite.
