increasing order of asymptotic complexity of functions $f_1, f_2, f_3$ and $f_4$ Question

Which of the given options provides the increasing order of asymptotic complexity of functions $f_1, f_2, f_3$ and $f_4$?

$f_1(n) = 2^n$
$f_2(n) = n^{3/2}$
$f_3(n) = n \log_2 n$
$f_4(n) = n^{\log_2 n}$
A.$f_3,  f_2, f_4, f_1$ 
B.$f_3, f_2, f_1, f_4$
C.$f_2, f_3, f_1, f_4$
D.$f_2, f_3, f_4, f_1$
This question i found in one of the entrance examination.I tried solving it but  i am getting wrong answer.so i want to verify that is my concept wrong or i am missing something?
Approach
I can write $f_1,f_2,f_3,f_4$ as

$$f_1=e^{n \log_2 2}$$
$$f_2=e^{\frac{3}{2} \log_2 n}$$
$$f_3=e^{ \log (n \log_2 n)}=e^{ \log (n )+\log \log_2 n}$$
$$f_4=e^{\log _2 n \times \log_2 n}$$

So, i am getting sequence as
$f_2, f_3, f_4, f_1$
But answer is given as-:
$f_3,  f_2, f_4, f_1$
I am not getting how

$$\frac{3}{2} \log_2 n > \log n +\log \log_2 n$$

 A: Basically, we are trying to calculate the limit $$\lim_{n\to\infty} \frac{n^{3/2}}{n\lg n}$$
(Note that $\log_2(n)$ and $\lg(n$ are asymptotically equavilent.)
Here is my derivation:
$$
\begin{aligned}
\lim_{n\to\infty} \frac{n^{3/2}}{n\lg n}
&= \lim_{n\to\infty}\frac{n^{1/2}}{\lg n} &\text{cancelling out n}\\
&= \lim_{n\to\infty}\frac{\frac{1}{2}n^{-1/2}}{\frac{1}{n}} &\text{L'Hospital's Rule}\\
&= \lim_{n\to\infty}\frac{1}{2}n^{1/2}\\
&= \infty
\end{aligned}
$$
Hence, $f_3<f_2$. Hope this helps!
A: $f_1(n)=2^n$
$f_2(n)=n^{3/2}$
$f_3(n)=nlog_2n$
$f_4(n)=n^{log_2n}$
Proof without using Calculus: 
$f_3=O(f_2):$
$nlog_2n\le n^{3/2}\Longleftrightarrow 2^{nlog_2n}=(2^{log_2n})^n=n^n\le 2^{n^{3/2}}=2^{nn^{1/2}}=(2^{n^{1/2}})^n=(2^{\sqrt{n}})^n$
$\Longleftrightarrow n\le 2^{\sqrt{n}}$
$n\le 2^{\sqrt{n}}:$
Will use following identity: $x^2\le 2^x$,$\space x\ge 4\space$(easily proven by induction). 
So if $n\gt 2^{\sqrt{n}}$,$\space(\sqrt{n})^2=n\le 2^{\sqrt{n}}\lt n$.  Contradiction.
Hence $nlog_2n\le n^{3/2}, n\ge 16$ 
$f_2=O(f_4):$
$n^{3/2}\le n^{log_2n}\Longleftrightarrow log_n{n^{3/2}}={3/2}\le log_n{n^{log_2{n}}}=log_2{n}\Longleftrightarrow 2^{3/2}\le n$
Hence $n^{3/2}\le n^{log_2n}, n\ge 2^{3/2}$
$f_4=O(f_1):$
$n^{log_2 n}\le 2^n\Longleftrightarrow log_2nlog_2n\le n\Longleftrightarrow log_2n\le\sqrt{n}\Longleftrightarrow 2^{log_2 n}=n\le 2^{\sqrt n}$ which has already been proven for $n\ge 16$. 
