Is solution unique for $\left(1+\frac{1}{x}\right)^x=c$ ($c$ is a positive number) when x can be negative? A youtube video suggests an equation $\left(1+\frac{1}{x}\right)^{x+1}=\left(1+\frac{1}{7}\right)^7$ and solves it in the following way:
$$
1+\frac{1}{x} = \frac{x+1}{x} = \left(\frac{x}{x+1}\right)^{-1}=\left(1-\frac{1}{x+1}\right)^{-1}
$$
Therefore:
$$
\left(1+\frac{1}{x}\right)^{x+1} = \left(1-\frac{1}{x+1}\right)^{-(x+1)}\
$$
Assuming $t=-(x+1)$ gives:
$$
\left(1+\frac{1}{t}\right)^{t}=\left(1+\frac{1}{x}\right)^{x+1}=\left(1+\frac{1}{7}\right)^7
$$
Hence $t=7$ and $x=-8$.
However can we really assume that $t=7$, given that $t$ may be either positive or negative?
Even if we take for granted that function $f(t)=\left(1+\frac{1}{t}\right)^t$ is monotonically increasing (which isn't really obvious), it isn't continuous being undefined for $t\in[-1,0]$ (where $1+\frac1t$ is undefined or non-positive), therefore I don't see a reason to claim that the solution is unique.
 A: Yes, I've also come to the same idea, as desdichado.
Take $\lim_{x\to -\infty}(1+\frac{1}{x})^x$ and make substitution $t=\frac{1}{x}$. So
$$
\lim_{x\to -\infty}\left(1+\frac{1}{x}\right)^x = \lim_{t\to -0}{(1+t)^\frac{1}{t}}= e^{\lim_{t\to -0}\frac{\ln(1+t)}{t}}
$$
Then by l'Hôpital's rule:
$$
\lim_{t \to -0}\frac{\ln(1+t)}{t}=\lim_{t\to-0}\frac{1}{t+1}=1
$$
Therefore $\lim_{x\to -\infty}\left(1+\frac{1}{x}\right)^x = e$. Given that function $\left(1+\frac{1}{x}\right)^x$ is increasing and continuous for $x \in (-\infty, -1)$ , it follows that $\left(1+\frac{1}{x}\right)^x>e$ for any $x \in (-\infty, -1)$.
A: After comments and your own answer, just for uour curiosity, I give below the only explicit solution for the equation
$$\left(1+\frac{1}{x}\right)^x=c$$ Let $x=\frac 1t$ to make
$$(1+t)^{\frac{1}{t}}=c\quad \implies \quad t =-\frac{W\left(-\frac{\log (c)}{c}\right)}{\log (c)}-1\quad \implies \quad x=$$ where $W(.)$ is Lambert function.
In the real domain, the argument must be  $\geq  -\frac 1e$, which, for this case, implies $c\geq e$.
