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For $n\ge 1$ and $a,b$ in a commutative ring, multiplying out the displayed polynomial factorization cancels all intermediate terms, leaving $a^n-b^n$. The case $n=2$ is the difference of squares. Unlike division by $a-b$, this identity also applies when $a=b$.

For real $1\le b\le a$, every summand lies between one and $a^{n-1}$. Consequently $n(a-b)\le a^n-b^n\le n(a-b)a^{n-1}$. This turns a one-step discrepancy into a bound after repeated multiplication, for example when estimating the global error of a one-step method on $y^{\prime }=y$.

This explicit numerical method is a two-stage Runge-Kutta method and a one-step method. For a sufficiently smooth ordinary differential equation, the chain rule gives $y^{\prime \prime }=f_x+f_{y}f$, and Taylor's theorem shows that its local truncation error is $O(h^3)$. Uniform Lipschitz continuity of $f$ in its solution argument gives an increment-function Lipschitz constant at most $L(1+hL/2)$. The Lipschitz one-step error bound then proves second-order numerical convergence on a fixed time interval. This method uses an intermediate prediction from $y_n$; the explicit midpoint linear multistep method instead requires a separate previous value $y_{n-1}$.

Suppose the increment function of a one-step method is Lipschitz continuous in its solution argument with constant $K>0$, uniformly in time and step size. Let $e_n=y(x_n)-y_n$ be the global error, and let $h{T}_n$ be the exact-start local truncation error. The triangle inequality gives $|e_{n+1}|\le (1+hK)|e_n|+h|T_n|$. Iterating and summing a geometric series proves the displayed estimate. Since $(1+hK{)}^n\le e^{Knh}$, uniformly bounded $K$ and consistency error $O(h^p)$ give global error $O(h^p)$ on a fixed time interval when $e_0=O(h^p)$. Here $O$ denotes Big O notation. This is a constant-step form of the discrete Gronwall inequality.

Write the increment function of the one-step method as $\Phi (x,v,h)=f(x+\beta h,v+\beta hf(x,v))$. Its consistency error isAs $h\to 0$, the first term tends to $y^{\prime }(x_n)$ and the second tends to $f(x_n,y(x_n))=y^{\prime }(x_n)$ by continuity. Both limits are uniform along the exact-solution curve, a compact set: its small stage perturbations lie in a fixed compact neighborhood, so uniform continuity applies there. Thus ${\max }_{n<N}|T_n|\to 0$, proving consistency for every $\beta \in [0,1]$.