I've shown $\lim_{n\to\infty}\sqrt[n]{n}=e$, but it should be $1$. Where's the mistake? $$\lim_{n\to \infty} \sqrt[n] n = \lim_{n\to \infty} n^{\frac{1}{n}} = \lim_{n \to \infty} \{(1+(n-1))^{\frac{1}{n-1}}\}{^{(n-1)\frac{1}{n}}} = \lim_{n\to \infty} e^{\frac{n-1}{n}} = e$$
But this clearly isn't true as the actual limit is $1$.
Where did I go wrong? 

$$\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=e$$
not
$$\lim_{u\to\infty}\left(1+u\right)^{1/u}= e$$
Let $A=\lim_{u\to\infty}\left(1+u\right)^{1/u}\implies\ln A=\lim_{u\to\infty}\dfrac{\ln(1+u)}u=\lim_{u\to\infty}\dfrac1{1+u}=0$ (Using L'Hôpital's Rule)
$\implies A=e^0$
