How to generalize the principle of mathematical induction for proving statements about more than one natural number? Suppose that $P(n_1, n_2, \ldots, n_N)$ be a proposition function involving $N >1$ positive integral variables $n_1, n_2, \ldots, n_N$. Then how to generalise the familiar induction to prove this statement? 
Can the following possibly be correct? 
(1) Suppose that $P(1, 1, \ldots, 1)$ is true. 
(2) Suppose the truth of $P(k_1, k_2, \ldots, k_N)$ implies the truth of $P(k_1+1, k_2+1, \ldots, K_N+1)$. 
Then $P(n_1, n_2, \ldots, n_N)$ is true for all $(n_1, n_2, \ldots, n_N) \in \mathbb{N}^N$. 
For example, can we generalise induction to prove the following statement? 

Let $m$ and $n$ be natural numbers such that $m < n$, and let $x$ be a real number such that $0 < x \leq \frac 1 2$. Then $$n^x - m^x < (n-m)x.$$ 

 A: While this can indeed be done, your described approach certainly isn't enough - here's a counterexample! Let $P(x_1, . . . , x_n)$ be the statement $$"x_1=x_2=...=x_n."$$ Then $P(1, . . . , 1)$ is true, and $P(x_1, . . . , x_n)$ implies $P(x_1+1, . . . , x_n+1)$, but clearly $P$ is not true in general.
What's going on is that this scheme fails to take into account the fact that we need to allow the variables to increment independently. Based on this, can you see what type of statement you'll need to prove in general?

If you get stuck, you might also google the phrase "double induction" and see if you can generalize that to $n$-many parameters. But I suggest trying to figure it out yourself, first.
A: The question about generalized induction has been dealt with in the answer by Noah Schweber. Here is a quick non-induction proof of the desired inequality. Note that it works under conditions substantially more general than the conditions in the OP.
For fixed exponent $x$, consider the function $f(t)=t^x$. Then by the Mean Value Theorem we have
$$\frac{f(n)-f(m)}{n-m}=f'(c)$$ for some $c$ between $m$ and $n$. 
Note that $f'(t)=xt^{x-1}$, so $f'(c)=xc^{x-1}$. But since $x-1$ is negative and $c\gt 1$, we have $c^{x-1}\lt 1$, and therefore $xc^{x-1}\lt x$. This completes the proof.
