The exponent isn't lost.  You must be confusing (x+5) with g(x+5). But 
g(x+5) stands for (x+5)^2, which has the exponent 2.

Maybe the following will help you understand.

Start with the function g(x) = x^2 whose graph looks like this:



Replace the x in g(x) = x^2 by (x+5) which gives you:

     g(x+5) = (x+5)^2

and that shifts the graph of g(x) to the left by 5 units, to look like this:



Now let's add 1 to both sides 

g(x+5) + 1  = (x+5)^2 + 1

which shifts the graph up 1 unit to look like this:



Now all that's been done now is to give this function a brand new name, h(x)

So you can either write it

h(x) = g(x+5) + 1

or you can write it as

h(x) = (x+5)^2 + 1 

which is the final way to write it

See how these last two equations are the same, since g(x+5) stands for
(h+5)^2?  Do you see that no exponent was lost? 

Edwin