Let $k^4\ge n < (k+1)^4$ for some positive integer $k$. Then, let $n=k^4+b$. Then, $\sqrt[4]{n} - [\sqrt[4]{n}]=\sqrt[4]{k^4+b} - k$. Notice that $0<\sqrt[4]{k^4+b}-k$ clearly for all $k$ and $b$. Also notice that $0<\sqrt[4]{k^4+b}-k$ for fixed $k$ is clearly minimized when $b=1$. Since the smallest $n$ would correspond to the smallest $k$, we set $b=1$ to find the smallest $k$ and so to find the smallest $n$. Then, $$\sqrt[4]{k^4+1}-k < \frac{1}{10^5}$$$$\iff \sqrt[4]{k^4+1} < k+ \frac{1}{10^{5}}$$$$\iff k^4+1 < k^4 + 4\cdot k^3 \cdot \frac{1}{10^5} + 6\cdot k^2 \cdot \frac{1}{10^{10}} + 4\cdot k \cdot \frac{1}{10^{15}} + \frac{1}{10^{20}}$$$$\iff 1 < 4\cdot k^3\cdot \frac{1}{10^5} + 6\cdot k^2 \cdot \frac{1}{10^{10}} + 4\cdot k \cdot \frac{1}{10^{15}} + \frac{1}{10^{20}}$$ Clearly, all the terms other than $4\cdot k^3\cdot \frac{1}{10^5}$ are very small in comparison especially for relatively small $k$ so we can just focus on the leading term. Then, $1< 4\cdot k^3\cdot \frac{1}{10^5} \iff \frac{10^5}{4} < k^3 \iff 2500 < k^3 \iff 14 \ge k$. So, the minimum $n$ is $14^3+1=2744+1=\boxed{2745}$