$\begin{align*}
&f1(n) = n^{0.999999} * \log n \approx n * \log n\\
&f4(n) = n^2 \\
&f3(n) = 1.000001^n \\
\\
&\text{we will show that } f_3 \gg f_4 \gg f_1 \\ \\
\hline \\
&1.\quad n^a \gg n\log n \qquad \text{where a > 1 and } n \rightarrow \infty \\ \\
&\lim_{n\rightarrow \infty}\left [ \frac{n\log n}{n^a} \right ] \qquad a > 1 \\
&=\lim_{n\rightarrow \infty}\left [ \frac{\log n}{n^{a-1}} \right ] \\
&=\lim_{n\rightarrow \infty}\left [ \frac{\frac{1}{n}}{(a-1)n^{a-2}} \right ] \\
&=0 \\
&\Rightarrow \text{This proves }f_4 \gg f_1 \qquad \text{for a = 2}\\
\\
\\
&\text{Now}, \\
&2. \qquad b^n \gg n^a \qquad \text{where a,b > 1 and } n \rightarrow \infty \\
\\
&\Rightarrow \lim_{n\rightarrow \infty}\left [ \frac{n^{k}}{b^n} \right ] \text{we will check this for k = 1} \\
\\
&\lim_{n\rightarrow \infty}\left [ \frac{n^{1}}{b^n} \right ] \\
&=\lim_{n\rightarrow \infty}\left [ \frac{n^{1}}{e^{n \ln b}} \right ] \\
&=\lim_{n\rightarrow \infty}\left [ \frac{1}{e^{n \ln b}.\ln b} \right ] \\
&=0 \qquad \left ( b > 1 \right )\\
\\
&\text{ Now, we will assume that for } k = a > 1 \;\; \lim_{n\rightarrow \infty}\left [ \frac{n^{k}}{b^n} \right ] = 0 \;\; \text{ holds}
\\
&\lim_{n\rightarrow \infty}\left [ \frac{n^{a+1}}{b^n} \right ] \\
&=\lim_{n\rightarrow \infty}\left [ \frac{n^{a+1}}{e^{n \ln b}} \right ] \\
&=\lim_{n\rightarrow \infty}\left [ \frac{(a+1).n^{a}}{e^{n \ln b}.\ln b} \right ] \\
&=\lim_{n\rightarrow \infty}\left [ \frac{(a+1).{\color{red}{n^a}}}{{\color{red}{b^n}}.\ln b} \right ] \\
&=\lim_{n\rightarrow \infty}\left [ {\color{red}{0}}.\frac{(a+1)}{\ln b} \right ] \\
&=0 \\ \\
&\text{Thus, the induction proves } \;\; b^n \gg n^a \qquad \text{where a,b > 1 and } n \rightarrow \infty \\
&\Rightarrow f_3 \gg f_4 \qquad \text{for large } \;\; n
\end{align*}$