More generally we take the degree to be $2n$. Let $P_\varepsilon (x)=x^{2n}+x+\varepsilon$. We note that $g:\mathbb{R}^+\rightarrow \mathbb{N}$ given by $g(\varepsilon)=f(P_\varepsilon)$ is "continuous" in the sense that if $x<y$ $g(x)=m>n=g(y)$ ($g$ is decreasing) and $m-n\geq 2$ then there exists $x<z<y$ such that $m>g(z)>n$. We then note that $g(1)=1$ because the minimum of $x^{2n}+x$ is strictly greater than $-1$, and $\lim_{\varepsilon\rightarrow 0} g(\varepsilon)=+\infty$ because $P_0(x)$ has a fixed point in 0 which never gets greater than 0. Thus by "MVT" (which is valid for this kind of "continuous" function)for each positive integer $k$ there exists an $\varepsilon$ such that $f(P_\varepsilon)=k$, and in particular if $2n=2014^{2015}$ and $k=2015$ we are done.