Why are you assuming that you can can just look at the the expression inside the parenthesis first? That is not valid because you are supposed to simultaneously keep raising it to higher and higher powers. The guy inside the parentheses is getting closer to 1, but it's not clear that that is pulling it down as fast as raising it to the nth power is pulling it up.
One way to understand what's going on more clearly is to use Euler's method.
This is a way to solve differential equations numerically.
In this case, what you want is a function that is its own derivative, so that means you are looking for a solution of the differential equation
\frac{dx}{dt} = x
subject to the initial condition
x(0) = 1
Geometrically, this means that you want a function whose slope is equal to its height. How can we find such a function? One approach is to imagine a "slope field". Set up coordinates on a plane so that x is the vertical direction and t is the horizontal. Now, for each point, we want a line whose slope is the x coordinate (remember, slope = height). We start out at the point (0,1). Then we go up the slope at that point for some time t/n. Then, when that time has passed, we end up at some other point. Now, we go up the slope at this new point. This procedure gives you the formula
(1+t/n)^n
As n goes to infinity, we should get a function that is closer and closer to what we want--its slope will be equal to its height. That's e^t.
It's clear that this thing is increasing, not staying equal to 1.
To get e, set t = 1 and you get your formula.
Draw a picture and work it out.
This construction is a special case of Euler's method.
http://en.wikipedia.org/wiki/Euler_method
Another way to visualize it is in terms of a vector field, rather than a slope field. In this case, you have a vector field on the real number line. The length of the vectors is proportional to the distance from 0. You start at 1 at time 0. So e^t is obtained by flowing along this vector field. And you could do the same thing we did with the slope field to get an approximation.
Yet another way to think of it is in terms of interest rates compounded at smaller and smaller time intervals, which gives the same formula, and also makes it clear that the thing is increasing, so that e has to be greater than 1.