how to prove $\left(\frac{n}{3}\right)^n\leq\frac{1}{3}n!$ i am asked to prove this statement:

$$\left(\frac{n}{3}\right)^n\leq\frac{1}{3}n!$$ 

Now after several attempts, i am lost not knowing where and how to start. if I use induction, i am stuck on the way. how can i solve this in an easy way? i have huge difficulties when i have to find something which is bigger than the latter term or so, because i lack some important source of maths in my brain. i am very likely to give up on the way if it becomes tough. for any guidance how to overcome this phase i will be very very thankful. i am trying for proficiency in math. Thanks

$$\left(\frac{n}{3}\right)^n\leq\frac{n!}{3}\Longleftrightarrow X_n:=\frac{n^n}{3^{n-1}n!}\leq 1$$
But
$$\frac{X_{n+1}}{X_n}=\frac{(n+1)^{n+1}}{3^n(n+1)!}\cdot\frac{3^{n-1}n!}{n^n}=\left(1+\frac{1}{n}\right)^n\frac{1}{3}\xrightarrow [n\to\infty]{}\frac{e}{3}<1$$
So that the infinite positive series $\,\displaystyle{\sum_{n=1}^\infty X_n}\,$ converges and thus $\,X_n\xrightarrow[n\to\infty]{}0\Longrightarrow X_n<1\,$ , at least 
from a certain index $\,n\,$ on.
