Why is O(n) better than O( nlog(n) )?

It turned out, I misunderstood Logn to be lesser than 1. As I asked few of my seniors i got to know this today itself, that if the value of n is large, (which it usually is, when we are considering Big O ie worst case), logn can be greater than 1.

So yeah, O(1) < O(logn) < O(n) < O(nlogn) holds true.

(I thought this to be a dumb question, and was about to delete it as well, but then realised, no question is dumb question and there might be others who get this confusion so I left it here.)

...because log n is always less than 1.

This is a faulty premise. In fact, log_b n > 1 for all n > b. For example, log₂ 32 = 5.

Colloquially, you can think of log n as the number of digits in n. If n is an 8-digit number then log n ≈ 8. Logarithms are usually bigger than 1 for most values of n, because most numbers have multiple digits.

Plot both the graph( on desmos (https://www.desmos.com/calculator) or any other web) and look yourself the result on large values of n ( y=f(n)). I am saying that you should look for large value because for small value of n the program will not have time issue. For convenience I had attached a graph below you can try for other base of log.

The red represent time = n and blue represent time = nlog(n).