 <blockquote> <p>Just wondering can i assume that 2^(loglogn) has the same growth as 2^n?</p> </blockquote> <p>No. Assuming that the logs are in base 2 then <code>2^log(n)</code> mathematically equals to <code>n</code>, so <code>2^(log(log(n)) = log(n)</code>. And of course it does not have the same growth as <code>2^n</code>.</p> <blockquote> <p>Should I take 1.1^n as a constant?</p> </blockquote> <p>No. <code>1.1^n</code> is still an exponent by <code>n</code> that cannot be ignored - of course it is not a constant.</p> <p>The correct order is:</p> <pre><code>2^loglogn = log(n) n,3n nlogn, 6nlogn 4n^2 7n^3 – 10n n^8621909 1.1^n n! </code></pre> <p>I suggest to take a look at the <a href="https://en.wikipedia.org/wiki/Big_O_notation" rel="nofollow">Formal definition of the Big-O notation</a>. But, for simplicity, the top of the list goes slower to infinity than the lower functions.<br> If for example, you will put two of those function on a graph you will see that one function will eventually pass the other one and will go faster to infinity.</p> <p>Take a look at <a href="https://www.wolframalpha.com/input/?i=n%5E2+and+2%5En" rel="nofollow">this</a> comparing <code>n^2</code> with <code>2^n</code>. You will notice that <code>2^n</code> is passing <code>n^2</code> and going faster to infinity.<br> You might also want to check the graph in <a href="http://bigocheatsheet.com/" rel="nofollow">this page</a>.</p>