Confusing moments about the Law of exponents for natural powers

Suppose that $$a\in \mathbb{R}$$ and $$a>1$$ and we define in the recursive way the following operation: $$a^1:=a,$$ $$a^{n+1}:=a^n\cdot a \quad\text{for each} \quad n\in \mathbb{N}.$$

I want to prove that $$a^{n+m}=a^na^m$$ for any $$n,m\in \mathbb{N}$$.

Proof: Let $$E=\{m\in \mathbb{N}: a^{n+m}=a^na^m \ \text{for any} \ n\in \mathbb{N}\}$$

For any $$n\in \mathbb{N}$$ we have $$a^{n+1}=a^na=a^na^1$$ which shows us that $$1\in E$$.

Suppose that $$m\in E$$ and we'll try to show that $$m+1\in E$$.

Indeed, for any $$n\in \mathbb{N}$$ we have $$a^na^{m+1}=a^n(a^ma)=(a^na^m)a=a^{n+m}a=a^{(n+m)+1}$$.

As you see I am almost done but one probably stupid moment bothers me a lot. Probably I am too pedantic.

We know that $$(n+m)+1=n+m+1$$ by associativity of natural numbers. How it follows that $$a^{(n+m)+1}=a^{n+m+1} \ ?$$

I do not think the we can just substitute $$(n+m)+1$$ by $$n+m+1$$.

May be I cannot formulate my question quite explicitly but I'd be very thankful for your help!

• You won't resolve your confusion if you write $n+m+1$. You surely mean $n+(m+1)$. – ancientmathematician Jan 24 at 12:18
• @ancientmathematician, could you give more details please? I will appreciate your help! – ZFR Jan 24 at 12:20
• That then reduces the question to proving that $+$ is associative on $\mathbb{N}$; which may depend on how you are defining $+$. – ancientmathematician Jan 24 at 12:20
• You get to $a^{(n+m)+1}$ OK; then this equals $a^{n+(m+1)}$ by associativity of $+$ on $\mathbb{N}$, so $m+1\in E$. – ancientmathematician Jan 24 at 12:22
• The definition you have given of $a^n$ is recursive: you give a base value then define $a^{n+1}$ in terms of $a^n$. It is true but non-trivial that there is one unique function that satisfies these conditions. And it is a function, that is each $n$ is related to precisely one $a^n$. [And no, this is not "injective" - quite the opposite in fact.] – ancientmathematician Jan 24 at 13:27

Yes, this last step is fine. If $$x=y$$ then $$a^x=a^y$$ and in this case you know that $$(n+m)+1=n+m+1$$.
I think you answered your own question. Because of associativity, $$(n+m)+1=n+(m+1)$$, so using your calculation, $$a^na^{m+1}=\cdots=a^{(n+m)+1}=a^{n+(m+1)}$$, so the inductive step is done.
Recall that associativity means $$(a+b)+c=a+(b+c)$$.
Another way of formulating it is that we have a function $$exp: \mathbb R \times \mathbb N \rightarrow \mathbb R$$ such that $$exp(a,n+1) = n* exp(a,n)$$ (in other words, I am rewriting $$a^n$$ as $$exp(a,n)$$). You are asking whether we can take $$exp(a,(n+m)+1)$$ to be the same as $$exp(a,n+m+1)$$. A fundamental property of a well-defined function is that if $$x$$ and $$y$$ are different expressions representing the same number, then $$f(x)=f(y)$$. So, for instance, $$f_1(\frac a b) = \frac {2a}b$$ is a well-defined function, because applying it to two different representation of the same fraction gives the same result (e.g. $$f_1(\frac 1 2)=f_1(\frac 2 4)$$). $$f_2(\frac a b) = a+b$$, on the other hand, is not a well-defined function, because $$f_2(\frac 1 2)=3$$, while $$f_2(\frac 2 4)=6$$.
You've said that you've defined $$exp$$ recursively, and implicit in saying that this is a definition is saying that the function is well-defined.