We will use the following well known fact: If $f(x)$ is a differentiable function in interval $(a,b)$ and $f(x)=0$ has $k$ roots (counting multiplicity) in $(a,b)$ then $f'(x)=0$ has at least $k-1$ roots in $(a,b)$. Now we write the given polynomial in the form $P(x) = 1+ x^n Q(x)$ where $\deg Q(x) = m \geq 0$ and $Q(x) $ is not identical with zero. Let on the contrary assume that $P(x)$ has at most $n-3$ non-real roots. Then $P(x)$ will have at least $m+n-(n-3) = m+3$ real zeroes. Let $k$ of them are $< 0$ and $m+3-k$ are $> 0$. (notice that 0 is not a root of $P(x)=0$ ). Thus $P'(x)$ will have at least $k-1$ roots in $(-\infty, 0)$ and at least $m+3-k-1$ roots in $(0, \infty)$, i.e. $P'(x)=0$ will have at least $m+1$ roots different to $0$. But $P'(x) = nx^{n-1} Q(x) + x^n Q'(x) = x^{n-1} (nQ(x)+xQ'(x)) \, \Rightarrow$ $nQ(x)+xQ'(x)$ has at least $m+1$ roots, which means that it is identical to zero. Hence $P(x)=const$, contradiction.