To compute $\lim_n (1+n)^{\frac1{\ln n}}$ without L'Hospital

We suppose that have this limit of a succession $n\in \Bbb N$ with real terms $$\lim_n (1+n)^{\frac1{\ln n}}$$ Now this limit have the value $e$. In fact I have done these steps remebering that $f^g=e^{g\ln f}$

$$\lim_n (1+n)^{\frac1{\ln n}}=\lim_n e^{ \frac{\ln (1+n)}{\ln n}}$$

Without L'Hospital's rule, I do not remember the result of $$\lim_n \frac{\ln (1+n)}{\ln n}$$ (related here - I have seen this now Limit of $\underset{n\to \infty }{\text{lim}}\frac{\ln (n+1)}{\ln (n)}$ without L'Hôpital) hence I had due to compute the $$\frac{\ln (1+n)}{\ln n}$$ when $n$ is more larger than 1. In fact if I choose $n=1000$, for example, I will have $\frac{\ln (1+n)}{\ln n} \to 1$ and $$\lim_n (1+n)^{\frac1{\ln n}}=\lim_n e^{ \frac{\ln (1+n)}{\ln n}}=e^1=e.\,\square$$

This way I do not like very much; hence I have tried another way to solve

$$\lim_n (1+n)^{\frac1{\ln n}}$$ being

$$\lim_n (1+n)^{\frac 1n}=e \tag 1$$

Thus, if I remember well, $\lim\limits_n \frac n{\ln n}=+\infty$ and $\lim\limits_n \frac{\ln n}n=0$, and

$$\lim_n (1+n)^{\frac1{\ln n}}=\lim_n \left[(1+n)^{\frac1{n}}\right]^{\frac n{\ln n}}=e^\infty=+\infty\neq e.$$ Actually I do not see my mistake. Thank you very much everyone.

Your mistake is that $\lim_{x\to0}(1+x)^{1/x}=e$, but $\lim_{n\to\infty}(1+n)^{1/n}\to1$. So the limit ends up not being $e^\infty$, but of the form $1^\infty$, which is indeterminate.

I always treat limits of this form by computing the limit of the logarithm, so we want to consider $$ \lim_{n\to\infty}\frac{\ln(1+n)}{\ln n} $$ If we consider instead of the sequence the function limit $$ \lim_{x\to\infty}\frac{\ln(1+x)}{\ln x} $$ we know that if this one exists, then it's the same of the limit of the sequence (the converse is generally not true). Now we can perform the substitution $x=1/t$ and note that $$ \ln(1+1/t)=\ln(t+1)-\ln t $$ so the limit becomes $$ \lim_{t\to0^+}\frac{\ln t-\ln(1+t)}{\ln t}=\lim_{t\to0^+}\Bigl(1-\frac{\ln(1+t)}{\ln t}\Bigr)=1 $$ because the fraction is $0/\infty$.

Thus we have proved that $$ \lim_{n\to\infty}(1+n)^{1/\ln n}=e^1=e $$