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I've just started learning limits and calculus. While solving a limit problem from my textbook, I found a different answer.

Problem: Find $\lim\limits_{n \to \infty} (1+\frac 1 n)^n$.
(The answer is obviously $e$, but I got a different answer.)
Solution from the book:
$\lim\limits_{n \to \infty} (1+\frac 1 n)^n=\lim\limits_{n \to \infty} \{1+\frac n {1!} \cdot \frac 1 n +\frac {n(n-1)}{2!}\cdot (\frac 1 n)^2+\frac{n(n-1)(n-2)}{3!}\cdot (\frac 1 n)^3+\dots \dots \}$
$=\lim\limits_{n \to \infty}\{1+\frac{1}{1!}+\frac{n^2(1-\frac 1 n)(1-\frac 2 n)}{2!} \cdot \frac 1 {n^2}+\frac{n^3(1-\frac 1 n)(1-\frac 2 n)}{3!}\cdot (\frac 1 n)^3+ \dots \dots \}$
$=\lim\limits_{n \to \infty}\{1+\frac{1}{1!}+\frac{(1-\frac 1 n)}{2!}+\frac{(1-\frac 1 n)(1-\frac 2 n)}{3!}+\dots \dots \}$
$=1+\frac 1 {1!} +\frac{(1-0)}{2!}+\frac{(1-0)(1-0)}{3!}+\dots \dots$
$=1+\frac 1 {1!} +\frac 1 {2!}+\frac 1 {3!}+ \dots \dots$
$=\color{green}{\boxed{e}}$
My solution:
We know that $\frac 1 n =0$ as $n$ approaches $+\infty$ ( this fact has also been used in the first solution) and $n$- th power of $1$ is $1$.
So, $\lim\limits_{n \to \infty} (1+\frac 1 n)^n=\lim\limits_{n \to \infty} (1+0)^n=\color{red}{\boxed{1}}$.

I don't find any mistake in my solution. So, what's wrong here?

Spectre
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Oshawott
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    The quantity inside is not EXACTLY 1, and as it approaches more towards 1, the power also increases. You are applying limits individually in the same term, which is strictly NO-NO while calculating limits. – bigbang Jun 28 '21 at 11:06
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    As $n$ increases, $1 + 1/n$ gets closer to $1$, but at the same time the exponent is growing larger. – littleO Jun 28 '21 at 11:25
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    By the same reasoning, $$\lim_{x \to \infty}x \cdot \frac{1}{x} = \lim_{x \to \infty}x \cdot 0=0 , .$$ – Joe Jun 28 '21 at 11:28
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    Is anyone else bothered by the book's "solution"? – David K Jun 28 '21 at 12:24
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    @DavidK - indeed. Although the ellipses inside the limit are actually finite since $n$ is integer. But a proper proof would show that the individual terms decrease the total with $n$ at least as fast as the new terms added increase the total. The problem can be corrected by noting that for each $n$, the limitend is bounded above by the first $n$ terms of the $e$ series. – Paul Sinclair Jun 28 '21 at 23:37

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