Why don't I get $e$ when I solve $\lim_{n\to \infty}(1 + \frac{1}{n})^n$? If I were given $\lim_{n\to \infty}(1 + \frac{1}{n})^n$, and asked to solve, I would do so as follows:
$$\lim_{n\to \infty}(1 + \frac{1}{n})^n$$
$$=(1 + \frac{1}{\infty})^\infty$$
$$=(1 + 0)^\infty$$
$$=1^\infty$$
$$=1$$
I'm aware that this limit is meant to equal to $e$, and so I ask:

why don't I get $e$ when I solve $\lim_{n\to \infty}(1 + \frac{1}{n})^n$?

 A: $1^{\infty}$ is an indeterminate form: the base is going to 1 which is trying to bring the expression to 1, but the exponent is going to $\infty$ which is trying to pull the expression off to $\infty$ as well.  That there are these opposing forces is how I try to explain why some forms of limits are indeterminate but other forms (like $0^{\infty}=0$) aren't.
A: I think it's instructive to look at what happens as n grows. At the bottom of my answer I've attached a plot showing $(1 + \frac{1}{n})^n$ as a function of n from 1 to 100.
Hopefully this plot makes a little intuitive sense. Although as $n\to\infty$, $(1 + \frac{1}{n})$ gets very small, you are raising it to a higher and higher power.
For example, when $n = 10$:
$$1 + \frac{1}{10} = 1.1$$
which is pretty close to $1$, but!
$$1.1^{10} \approx 2.6$$

To answer your question: You cannot substitute the limiting value in the expression and get the proper limit. As an example, consider
$$f(x) = \frac{(x^2 - 1)}{x-1}$$
If we try to take the limit as $x \to 1$ by simply substituting $x = 1$, we get
$$\lim_{x \to 1}f(x) = \frac{1^2 - 1}{1 - 1} = \frac{0}{0}$$
whereas the true limit can be found by factoring the numerator:
$$f(x) = \frac{(x-1)(x+1)}{x-1} = x + 1$$
so:
$$\lim_{x \to 1}f(x) = 2$$
See the Wikipedia page on limits.
A: The main problem here is that you use $\infty $ as if it were a number, and then you just substitute $\infty $ for $n$ and 'compute'. But, since infinity is not a number, you can't just substitute it and the computation you make is meaningless. 
I'm assuming from the way you ask the question that you already know how to derive the correct value and that you are just wondering what is wrong with your approach. So, you don't get $e$ when you compute the limit they way you did precisely because your computation is invalid since you treat infinity as a number. 
This may be extra confusing since sometimes substituting $\infty $ does lead to the correct answer, but this should be regarded as a fluke. For instance, the limit $\lim _{n\to \infty }\frac {1}{n}$ is $0$, which is what you would get by substituting $\infty $. But this is just coincidence that a completely faulty line of argument using entirely wrong 'computations' leads to the correct answer. Unfortunately, often when teaching calculus such substitutions and manipulations with $\infty $ are glossed over, or worse even encouraged. It is good practice to never ever substitute $\infty $ and compute with it.
A: This is actually a fairly good intuitive way to think about the problem but, as you point out, it does lead to the wrong answer.  The mistake you're making is to use intuition instead of recognised theorems to manipulate limits.  Get a textbook (or website) on limits, look up the results, and check that your assertions about the manipulations follow from these theorems. 
Besides your wrong answer of 1, very similar reasoning to yours can be used to obtain the answer of $\infty$. This false reasoning would work as follows.  If $x > 1$, then $x^\infty = \infty$.  Since $1 + \frac{1}{n}$ is always we greater than $1$, the limit is $(1 + \frac{1}{n})^\infty = \infty$ 
Unsurprisingly, the actual answer of $e$ lies between these two wrong answers.  To actually solve your problem, evaluate the complete expression of $(1 + \frac{1}{n})^n$.  Imagine that n is a large real number (like a million) and examine the expansion.  You should see that it is similar to the standard expression for $e$, and that should help you to prove formally that the expansion tends to $e$ as $n$ tends to $\infty$.
Paul Epstein
A: Consider $$\lim_{n\rightarrow \infty}(1+{1\over n})^n.$$ This limit has the indeterminate form $1^\infty$. Let $y=(1+{1\over n})^n$. Taking the natural logarithm of both sides of the equation and simplifying using the rules of logarithms we obtain $\ln(y)=n\ln(1+{1\over n})$. The $$\lim_{n\rightarrow \infty} \ln(y)=\lim_{n\rightarrow \infty}n\ln(1+{1\over n})$$ which has the indeterminate form $\infty\cdot 0$. We can rewrite the right-hand side limit as $$\lim_{n\rightarrow \infty}\ln(y)={\lim_{n\rightarrow \infty}={\ln(1+{1\over n})\over {1\over n}}}$$ which has the indeterminate form ${0\over 0}$. Using L'Hospital's Rule we see that $$\lim_{n\rightarrow \infty}\ln(y)=\lim_{n\rightarrow \infty}{{1\over (1+{1\over n})}\cdot {-1\over n^2}\over {-1\over n^2}}.$$ This simplifies to $$\lim_{n\rightarrow \infty} \ln(y)=\lim_{n\rightarrow \infty}(1+{1\over n})=1.$$ So far we have computed the limit of $\ln(y)$, what we really want is the limit of $y$. We know that $y=e^{\ln(y)}$. So $$\lim_{n\rightarrow \infty}(1+{1\over n})^n=\lim_{n\rightarrow \infty} y=\lim_{n\rightarrow \infty} e^{\ln(y)}=e^1=e.$$ Thus $$\lim_{n\rightarrow \infty} (1+{1\over n})^n=e.$$
A: Hint : $1^\infty=(a^0)^\infty=a^{0\cdot\infty}=a^\text{Indeterminate}=\text{Indeterminate}$.
A more comprehensive list of such undetermined expressions can be found here.
A: The problem is $1 + 1/n \to 1$ as $n \to \infty$, which would tend to make the limit close to or equal to $1$, but at the same time the exponent $n$ is approaching infinity, which would tend to make the limit very large.  It's not apparent which is more important, and whether the limit is $1$, $\infty$, some number in between, or maybe it doesn't exist at all.  It turns out that, roughly speaking, the convergence of $1 + 1/n$ to $1$ and the convergence of $n$ to $\infty$ are equally important, and the limit turns out to be something "in between", namely $e$.  Someone will probably post a complete proof that the limit is $e$, but that is the idea.  
A: Noting : $(1+1/n)^n=e^{n \log(1+1/n)}$
And when $x\to 0$: $\log(1+x) = x+o(x)$
Or again $n \to \infty$: $log(1+1/n)=1/n+o(1/n)$
Scope : $(1+1/n)^n=e^{1+o(1)}$
A: Answer: It is same thing to prove $(1+x)^{1/x}$ as $x$ goes to $0$ its limit is e.
Say $(1+x)^{1/x}=y$  then take log of both side. Then take limit as x approaches to 0 of both sides. Then Use L'Hopital. You will get exactly e. 
A: The rule of thumb with these "determinate" and "indeterminate" forms is if you can think of a way to screw it up, i.e., get a different answer depending on how you look at it, it's indeterminant. For example, $\frac{0}{0}$, the classic indeterminate form, is broken by $1 = \frac{x}{x} = \frac{0}{0}$ as $x\to 0$ and $x=\frac{x^2}{x}$ goes to $0$ as $x\to 0$. So we can equally interpret $\frac{0}{0}$ as $1$ or as $0$.  Put another way, it really depends on how fast the numerator and denominator approach $0$ relative to one another. 
The way to think of $1^\infty$ is this. $$x^\infty=\begin{cases}\infty,&\text{if }x > 1\\0,&\text{if }0 < x < 1\end{cases}$$ (more formally $\lim_{a\to\infty}x^a$ equals the right-hand side). So if you interpret $1^\infty$ using the first formula, you get $\infty$, and if you interpret in using the second, you get $0$. You can interpret it a third way as you saw as $1^a = 1$. What it boils down to is that $1^\infty$ really depends on how fast the base approaches one relative to how fast the exponent approaches $\infty$. You can get any number you want by varying the rate (or at least any nonnegative number), including $e$. 
A: Observe that
$$\left(\frac{2n+1}{2n}\right)^{2n}=\left(\frac{4n^2+2n+1}{4n^2}\right)^n>\left(\frac{4n^2+2n}{4n^2}\right)^n=\left(\frac{n+1}n\right)^n.$$
so that there is an increasing subsequence.
If the limit exists, it must exceed $2$ (and also $1.000000001^{1000000000}\approx2.7182818270999\cdots$).
