6. -2015 .

Notice that for any $x \in(-1,+\infty)$ we have
$$\frac{x}{1+x} \leqslant \ln (1+x) \leqslant x \text {. }$$

Then for $f(x)=\left(1+\frac{1}{x}\right)^{x+1}(x>0)$ and
$$g(x)=\left(1+\frac{1}{x}\right)^{x}(x>0)$$

the derivatives are respectively
$$\begin{array}{l} f^{\prime}(x)=\left(1+\frac{1}{x}\right)^{x+1}\left[\ln \left(1+\frac{1}{x}\right)-\frac{1}{x}\right]0 . \end{array}$$

Thus, $f(x)$ is decreasing on the interval $(0,+\infty)$, $g(x)$ is increasing on the interval $(0,+\infty)$, and for any $x \in(0,+\infty)$ we have $f(x)>\mathrm{e}>g(x)$.
Therefore, for any $m 、 n \in \mathbf{Z}_{+}$ we have
$$\left(1+\frac{1}{n}\right)^{n+1}>\mathrm{e}>\left(1+\frac{1}{m}\right)^{m} \text {. }$$

Thus, the integer $n$ that satisfies $\left(1+\frac{1}{n}\right)^{n+1}=\left(1+\frac{1}{2014}\right)^{2014}$ must be negative.
Let $n=-k\left(k \in \mathbf{Z}_{+}\right)$, substituting into the given equation we get
$$\left(1+\frac{1}{2014}\right)^{2014}=\left(1-\frac{1}{k}\right)^{-k+1}=\left(1+\frac{1}{k-1}\right)^{k-1} \text {. }$$

Hence $k-1=2014, n=-k=-2015$.