"If $n=0$ there is nothing to prove" Suppose that I must prove a theorem by (strong) induction on $\mathbb N$. If the statement of the theorem has sense, for example, for $n\ge 1$, often as base case one chooses $n=0$ and says "for $n=0$ there is nothing to prove". So instead to start proving the base case for $n=1$, the induction "begin" from a statement that formally doesn't have sense.
I don't understand why this procedure is allowed.
 A: The statement isn't, or rather shouldn't be, nonsensical - just trivial (the meaning of which varies on the author's intended audience, of course). That is, "nothing to prove" just means "we have no work to do because this fact is obvious".
For example, one of my professors who is a topologist says that, when he has the opportunity, he loves to start an inductive argument on the spheres $S^n$ by starting at $n=0$ (when the sphere $S^0$ is just the two-point discrete space), or even $n=-1$ when possible (when $S^{-1}$ is the empty set).
A: Induction is really just an application of the following theorem:

Let $P(x)$ be some property of $x$. If $P(0)$ is true, and for every $n\in\Bbb N$, $P(n)\rightarrow P(n+1)$; then for every natural number $P(n)$ holds.

That means that if a certain property holds for zero; and whenever it holds for $n$ we can prove it holds for $n+1$, then this property holds for every natural number. This, of course, saves us infinitely many schematic proofs. 
For example the case of $n<2^n$; we want to show it is always true. So theoretically we had to take every natural number separately and show this inequality holds for each case. But those are infinitely many proofs to write, and we only have so many trees to turn into papers, moreover we may die sometime within the first $2^{10000}$ cases. So instead we need to find a better method, such method would be to take an arbitrary natural number and to show this is true for that number. Since we took an arbitrary number, it must be true for any number we would take.
But not all proofs can be easily written for a completely arbitrary case, and often we find ourselves in the need for extra assumptions. Induction allows us to have that one extra assumption, that whatever we want to prove, was true for the previous number. Then, the induction process tells us, verify that zero has the wanted property and you're done!
Luckily for us, many claims that we prove by induction are vacuously, or trivially true for zero. For example, summation formulas are trivially true for zero because summing no elements is just zero.
Some people prefer to start from the least non-trivial case. It is sometimes instructive to do so (e.g. when giving the first examples of induction); and it may be that you are going to have to separate that case from the rest anyway. But it is also important to understand that even if we have nothing substantial to prove for the case $n=0$, it's fine if we start there.
A: Saying "there is nothing to prove for $n=0$" (say) should mean exactly that; the statement has a well defined meaning for $n=0$, but it is vacuously (not trivially, which is a subjective term) satisfied. For instance the statement may claim that all elements of some $n$-set have a certain property; for the empty set one cannot even start considering such an element. Or the hypotheses of the statement might be impossible to satisfy for $n=0$. Or other similar circumstances where one simply cannot find anything that needs consideration.
Since you mentioned the Jordan-Hölder theorem in a comment, I looked at that, and it is not really an example of "nothing to prove" for the case $|G|=1$. That group has a unique composition series, of length$~0$, and the result is true trivially, because two elements of a singleton set are necessarily equal, but not vacuously. It would be vacuously true for $|G|=0$ if that were taken as the starting case, since there aren't any groups with $0$ elements. Whether one can actually use that as starting case depends on the form of the inductive step however; if it uses that the series has positive length, then it cannot be used for 
$|G|=1$.
