$e^{H_n}$ is not an exponential function. Is the composition of an exponential function with the harmonic number.

You wouldn't call the function $f(x)=x$ exponential, in spite of having $f(x) = e^{\log x}$, right?

You'll find the proof that any exponential function $f(x) = a^x$ with $a>1$ grows faster than any polynomial function in almost any Calculus book. Or you can try proving it using L'Hopital's rule.

@jjagmath If e^H_n is not an exponential function, then what kind of function is it?

@jjagmath Well, I guess it is an example of a composite function, which is a function composed of two other functions. In this case, the two functions are the exponential function e^x and the harmonic number function H_n. Although it's not a simple exponential function, I think it is still an exponential function. What do you think?

I think it is still an exponential function. --- Would you call $x^2 - x(x-1)$ a quadratic function?

@Ok-Virus2237 So you do think $f(x) = e^{\log x}=x$ is an exponential function just because is the composition of an exponential function with other function?

If you are calling the function $f(x)=x$ exponential I won't continue arguing with you.

Well, I was only thinking, I am no expert and you being a PhD holder, I would appreciate and highly respect your opinion. So I was just asking. Now I got it, thanks.

Well, if you're asking then read the answers. Already in my second comment I gave you the example that refutes your statement.

@jjagmath Still if you can tell me what kind of function it is exactly? And what are its properties?

The function is almost linear. It can be proved that $e^{H_n} = a n + b + \text { some other terms that tend to $0$}$, where $a$ and $b$ are some constants.

Thanks. Just another perspective, does the two functions diverge as n increases, and e^H_n grows faster than e^n? Like sort of a transcendental function that involves special functions and constants.

Both tend to infinity as $n$ grows, and, as I said, the function $e^{H_n}$ is almost linear so it grows much slower than $e^n$.

@jjagmath But the reason I think so is because H_n is asymptotic to log(n) + gamma, where gamma is the Euler-Mascheroni constant. Therefore, e^H_n is asymptotic to e^(log(n) + gamma), which is equivalent to n * e^gamma. On the other hand, e^n is asymptotic to e^n. Since n * e^gamma grows faster than e^n as n increases, e^H_n grows faster than e^n as well. The two functions have the same value at n = 1, but then e^H_n starts to increase more rapidly than e^n. Is the logic wrong?

Why do you think $n e^\gamma$ grows faster than $e^n$?

I got the flaw now. Thank you, Sir, for pointing it out.

This particular example is suspiciously closely related to Lagarias' criterion for the Riemann Hypothesis, which states that the RH is equivalent to $\sigma(n) < H_n + e^{H_n}\log H_n$ for all $n\ge 1$. That should be a hint it is incredibily hard to prove.

Those who downvote for no reason should be banned from this site.

@AliShadhar Yes, I agree! The person who downvoted should atleast give proper reasons. It will help me get better next time. Also I don't think this question is useless enough to get downvoted.

Most of these downvoters have nothing to do with math, so don't expect they have the knowledge or courage to give a reason