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This explanation may work for some but not for others--it is perhaps a matter of taste or intuition whether you feel that it is sensible.  Is doing nothing a "way of arranging" zero objects?  Eh, you get get very philosophical about this and probably not in a productive way.  

I find the following argument more satisfying for why we should take 0!=1.  It is because 0! is itself not meaningful, so in a sense, we should not say that it equals anything.  In another sense, we are free to say that it equals whatever we want.  After all, we may define any symbol to be the name of any function.  So we can choose to take n! to name any function we want.  

For n>0 we take n! to be the number of arrangements of n objects.  We are not compelled to define n at 0.  But if we want to, here's a good reason to define it to be 1:  It makes writing other sensible equations easier.  For instance, take the sequence 

$$ 1, x, \frac{x^2}{2}, \frac{x^3}{3!}, \frac{x^4}{4!}, ... $$

In this sequence we keep increasing the degree of the monomial, and the factorial in the denominator.  This is an important sequence because we can use it to define $e^x$.  Notice that every term except the first can be described as 

$$ \frac{x^n}{n!} $$

If we define $n!=1$ then this equation also works for the first.

Here's another example:  Combinations.  Suppose that you want to count the number of ways to choose $k$ objects from a collection of $n$ objects.  It turns out that a formula for this is given by 

$$ \binom n k = \frac{n!}{k!(n-k)!} $$

and it makes sense for all $0<k<n$.  But what if you want to choose zero objects from $n$ objects?  Well there is just one way to do that:  Choose no objects!  

I think choosing no objects makes more sense than arranging zero objects, because even when you choose no objects from say 10 objects, those 10 objects are actually there.  So it makes sense to me that there is one way to not choose any of the objects that are available to you: Just stand there like an idiot.  

So hopefully this argument is convincing that we should have $\binom n 0 = 1$.  That is to say, for any appropriate $n$, the number of ways to not choose any is 1.  Well, if you take $0!=1$ then the same formula from before works.  Because with that formula you get 

$$ \binom n 0 = \frac{n!}{0!(n-0)!}=\frac{n!}{n!}=1 $$

and in fact it also makes sense when $k=n$.

$$ \binom n n = \frac{n!}{n!(n-n)!} = \frac{n!}{n!} = 1 $$

That is also, intuitively the number of ways of selecting all the objects.  

Many other examples are possible, where our formulas are simpler to write and continue to make sense, when we take $0!=1$.