Without using L'Hospital's rule and series expansion solve $e\lim\limits_{n \to \infty}\frac{e^{n\ln(1+1/n)-1}-1}{n\ln(1+1/n)-1}n(n\ln(1+1/n)-1)$

Question

This is my problem:

Solve only with the basic rules of natural logarithm and the definition for $e$.

$\lim\limits_{n\to\infty}(n(1+\frac{1}{n})^n - ne)$

And I got till here:

$e\lim\limits_{n \to \infty}\frac{e^{n\ln(1+\frac{1}{n})-1}-1}{n\ln(1+\frac{1}{n})-1}n(n\ln(1+\frac{1}{n})-1)$.

Both of these equations are equivalent with the limit $\frac{-e}{2}$. But I don't know how to transform to the next step. The first fraction should short itself giving me $1$ and the second multiplicand should give us $\frac{-1}{2}$ with the $e$ in front of the limit giving us the correct answer.

The definition for $e$ is $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n = e$.

But I think in this last step we should make a substitute $n = \frac{1}{x}$ and use the other definition $\lim\limits_{x\to0}(1+x)^{\frac{1}{x}} = e$

Answer 1

Let me write found inequalities here $$\frac{1}{2n^2}- \frac{1}{n^3}\leqslant \frac{1}{n}-\ln\left(1+ \frac{1}{n}\right)\leqslant \frac{1}{2n^2}$$

p.s. I obtained expression identical to OP in previous question, but keep answer deleted, as hadn't inequalities in that time. We can prove them by examining the corresponding functions, but maybe someone will be able to see an elementary proof.