Fundamental Limit Question I know that  $a_n=1+\frac1n$ converges to $1$. 
How do you prove that $b_n=\left(1+\frac{1}{n}\right)^c$ converges to $1$ where  $c\in   \mathbb{N}$ is a constant(unfortunaley originally I wrote $c\in\mathbb{R}$ which led to all the comments), and why doesn't the same proof work for $c_n=\left(1+\frac{1}{n}\right)^n$ which converge to $e$. 
Please try to prove it using elementary tools (such as the definition of limit, limit arithmetic ...).
Thank you very much.
 A: Among the limit laws, we have:

If $\lim\limits_{n\to\infty}a_n = r$ and $\lim\limits_{n\to\infty}b_n = s$, then $\lim\limits_{n\to\infty}a_nb_n = rs$.

Using this law, it is easy to check that for a fixed integer $k$, we have:

If $\lim\limits_{n\to\infty}a_n = r$, then $\lim\limits_{n\to \infty}a_n^k = r^k$.

This is straighforward: we are doing the same thing to each term of the sequence (raising to the $k$th power), so we expect to end up doing the same thing to the limit.
But if the exponent is variable, then we are doing different things to each term of the sequence. We take $a_1$ and raise it to the first power; $a_2$ is raised to the second power; $a_3$ to the third power; etc. We do different things to different terms. Consider for example the sequence $a_n=\frac{1}{n}$. The limit is $0$; if we multiply all terms by $10000$, then the limit is still $0$. But what if we multiply each term by something else? If we take $na_n = 1$, then the limit is $1$. If we take $2na_n$, then the limit is $2$. Etc. It doesn't make sense to say 

since $\lim\limits_{n\to\infty}a_n = 0$, then $\lim\limits_{n\to\infty}na_n=0$, because the $a_n$s go to $0$, and then multiplying by $n$ we get $0$ again.

any more than it would make sense to say

since $\lim\limits_{n\to\infty} n = \infty$, then $\lim\limits_{n\to\infty}na_n = \infty$ because the $n$s go to $\infty$, and then multiplying by $a_n\gt 0$ we get $\infty$ again.

Since both factors are changing, we need to analyze them together (the limit law about products doesn't work here because we get $0\times\infty$, which is not defined). The same is true with $(1+\frac{1}{n})^n$. You have a number, $1+\frac{1}{n}$, which is getting smaller, raised to an exponent which is getting larger. The two are "battling each other out". The base is trying to get the whole thing down to $1$, the exponent is trying to get the whole thing up to $\infty$. We cannot declare ahead of time that the base will win, any more than we can declare ahead of time that the exponent will win: this is a problem of whether the two impulses, one getting stronger and stronger towards $1$, and one getting stronger and stronger towards $\infty$, balance each other out (if so, where?) or if one of them beats the other (if so, which)? This is not the case in $(1+\frac{1}{n})^k$, because the exponent is not changing, so it exerts a "constant force" towards $\infty$, so the base, which is exerting stronger and stronger forces towards $1$, will eventually drown it out.
We can see, at least, that whatever $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n$ might be, it must be something greater than $1$. Note that by the binomial expansion, we have:
$$\begin{align*}
\left(1+\frac{1}{n}\right)^n &= 1 + \binom{n}{1}\frac{1}{n} + \binom{n}{2}\frac{1}{n^2}+\cdots + \binom{n}{n-1}\frac{1}{n^{n-1}} + \frac{1}{n^n}\\
&\gt 1 + \binom{n}{1}\frac{1}{n} \\
&= 1 + \frac{n!}{1!(n-1)!}\frac{1}{n}\\
&= 1 + \frac{n}{n}\\
&= 2.
\end{align*}$$
So, since each term of the sequence is greater than $2$, the limit, if it exists, is at least $2$. 
On the other hand, one can also show that the limit, if it exists, is no more than, say, $4$, though this is harder to do. So the two forces will likely balance each other out somewhere between $2$ and $4$ (and in fact they do).

Below is an answer that assumes knowledge of continuity.

One way to show that $b_n$ converges to one is to note that the function $f(x) = x^c$ is continuous at $1$ for every $c$. We have the following limit law:

Limit law. If $\lim\limits_{x\to c}g(x) = a$, and $f(x)$ is continuous at $a$, then $\lim\limits_{x\to c}f(g(x)) = f\left(\lim\limits_{x\to c}g(x)\right) = f(a)$.
Limit law with sequences. If $\lim\limits_{n\to\infty}d_n = a$ and $f(x)$ is continuous at $a$, then $\lim\limits_{n\to \infty}f(d_n) = f(a)$.

(In fact, one can define continuity of $f(x)$ at $a$ by saying that $f(x)$ is continuous at $a$ if and only if whenever $d_n$ converges to $a$, $f(d_n)$ converges to $f(a)$).
So for fixed $c$, we have $f(x)=x^c$ is continuous at $x=1$, and we have
$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^c = \lim_{n\to\infty}f\left(1+\frac{1}{n}\right) = f\left(\lim_{n\to\infty}1+\frac{1}{n}\right) = f(1) = 1^c = 1.$$
The same argument does not work for $(1+\frac{1}{n})^n$. Here we don't have a single function we want to "pass the limit through"; rather, we are changing both the input and the function. This is bound to lead to complications. We can turn this into a single function by remembering that
$$a^b = e^{b\ln a},$$
and rewriting as
$$\left(1+\frac{1}{n}\right)^n = e^{n\ln(1+\frac{1}{n})}.$$
Now, $e$ is continuous everywhere, so it is true that
$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = \lim_{n\to\infty}e^{n\ln(1+\frac{1}{n})} = e^{\lim\limits_{n\to\infty}n\ln(1+\frac{1}{n})}.$$
But now you'll notice that the limit you get in the exponent is an $\infty\times 0$ indeterminate, so that means you need to figure out how much it is. It's not equal to $0$, as it happens, which is why the limit is not equal to $1$.
(Also: if you could evaluate the limit of $(1+\frac{1}{n})^n$ by first taking the limit inside the parentheses and then the exponent, why not the other way? Then you would first take the limit of the exponent, which would give you $\infty$, and then the limit of $\infty$ is just $\infty$. The value of the limit should not depend on how we evaluate it, which suggests that taking the two expressions of $n$ and computing their limits separately in some order is not a valid way to evaluate this limit).
A: If $c \in \mathbb{N}$, the answer is really easy! You know that $$\lim_{n \to +\infty} p_n q_n = \left( \lim_{n \to +\infty} p_n \right) \left( \lim_{n \to +\infty} q_n \right)$$ whenever the right-hand side contains finite limits. Hence you deduce that $$\lim_{n \to +\infty} p_n^2 = \left( \lim_{n \to +\infty} p_n \right)^2,$$ and, by recurrence/induction, $$\lim_{n \to +\infty} p_n^c = \left( \lim_{n \to +\infty} p_n \right)^c$$ for any $c \in \mathbb{N}$.
