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I hope the following will satisfy Bjorn. It is a proof by induction which naturally skips over the base case and is at the undergraduate level. I saw the argument for the first time today in the paper containing 10 proofs in Russian of the Fundamental Theorem of Algebra which Ilya Nikokoshev made a link to in his answer to a question asking for lots of different proofs of that theorem. Here we go:

Claim: A nonzero polynomial (over a field) of nonzero degree has no more roots than its degree.

Proof: We prove this by induction on the degree $n$ of the polynomial. Assume that the polynomial $$p(X) = a_nX^n + \cdots + a_1X + a_0$$ of degree $n$ has at least $n+1$ different roots $r_1,\dots, r_{n+1}$. Consider the polynomial $$q(X) = a_n(X-r_1)\cdots (X-r_n).$$ We have $p(X) \not= q(X)$ since $p(r_{n+1}) = 0$ and $q(r_{n+1}) \not= 0$. The difference $d(X) = p(X) - q(X)$ is a nonzero polynomial of degree less than $n$ having at least $n$ roots $r_1,\dots,r_n$. This contradicts the inductive hypothesis. QED

[EDIT: IGNORE what follows in the next paragraph, which was in the original post, since I confused myself about strong vs. ordinary induction. The above proof is by strong induction since the degree of d(X) is merely less than n and not necessarily n-1 itself.]

One aspect of this which does not fit Bjorn's request is that this argument uses ordinary induction, not strong induction. But really, is that such a big deal? I suspect his main interest is seeing an inductive argument at all where the base case is naturally not mentioned, rather than specifically one using strong induction.

Proof: We prove this by induction on the degree $n$ of the polynomial. Assume that the polynomial $p(X)$p(X) = a_nX^n + \cdots + a_1X + x_0$a_0$$of degree n has at least n+1 different roots r_1,\dots, r_{n+1}. Consider the polynomial q(X) q(X) = a_n(X-r_1)\cdots (X-r_n). X-r_n).$$ We have$p(X) \not= q(X)$since$p(r_{n+1}) = 0$and$q(r_{n+1}) \not= 0$. The difference$d(X) = p(X) - q(X)$is a nonzero polynomial of degree less than$n$having at least$n$roots$r_1,\dots,r_n$. This contradicts the inductive hypothesis. QED [EDIT: IGNORE what follows in the next paragraph, which was in the original post, since I confused myself about strong vs. ordinary induction. The above proof is by strong induction since the degree of d(X) is merely less than n and not necessarily n-1 itself.] One aspect of this which does not fit Bjorn's request is that this argument uses ordinary induction, not strong induction. But really, is that such a big deal? I suspect his main interest is seeing an inductive argument at all where the base case is naturally not mentioned, rather than specifically one using strong induction. 2 added 260 characters in body; edited body I hope the following will mostly satisfy Bjorn. It is a proof by induction which naturally skips over the base case and is at the undergraduate level. I saw the argument for the first time today in the paper containing 10 proofs in Russian of the Fundamental Theorem of Algebra which Ilya Nikokoshev made a link to in his answer to a question asking for lots of different proofs of that theorem. Here we go: Claim: A polynomial (over a field) of nonzero degree has no more roots than its degree. Proof: We prove this by induction on the degree$n$of the polynomial. Assume that the polynomial$p(X) = a_nX^n + \cdots + a_1X + x_0$of degree$n$has at least$n+1$different roots$r_1,\dots, r_{n+1}$. Consider the polynomial$q(X) = a_n(X-r_1)\cdots (X-r_n)$. We have$p(X) \not= q(X)$since$p(r_{n+1}) = 0$and$q(r_{n+1}) \not= 0$. The difference$d(X) = p(X) - q(X)$is a nonzero polynomial of degree less than$n$having at least$n$roots$r_1,\dots,r_n\$. This contradicts the inductive hypothesis. QED