Global error bound for Euler's method

Write the local error committed in one step as the difference between the exact solution advanced by Taylor expansion and the Euler update. Taylor's theorem gives $y(t_{n+1}) = y(t_n) + h f(t_n, y(t_n)) + \frac{h^2}{2} y''(\xi_n)$ for some $\xi_n$ between $t_n$ and $t_{n+1}$, so the local truncation error is $\tau_n = \frac{h^2}{2} y''(\xi_n)$, bounded by $\frac{h^2 M}{2}$ using the assumption $|y''(t) | \le M$.

Subtract the Euler update $y_{n+1} = y_n + h f(t_n, y_n)$ from this exact Taylor expansion. Writing $e_n = y(t_n) - y_n$, the difference of the two right-hand sides gives $e_{n+1} = e_n + h\left[f(t_n, y(t_n)) - f(t_n, y_n)\right] + \tau_n$. The Lipschitz condition bounds the bracketed term by $L|e_n|$, so $|e_{n+1}| \le (1+hL)|e_n| + \frac{h^2 M}{2}$.

Since $e_0 = 0$ (the starting value is exact), unrolling this recursion by induction gives $|e_n| \le \frac{h^2M}{2}\sum_{k=0}^{n-1}(1+hL)^k = \frac{hM}{2L}\left[(1+hL)^n - 1\right]$, using the geometric-series identity. Finally, $(1+hL)^n \le e^{Lhn} = e^{L(t_n-t_0)}$ because $1+x\le e^x$ for all real $x$, which turns the bound into $|e_n| \le \frac{hM}{2L}\left(e^{L(t_n-t_0)}-1\right)$ — exactly the claimed inequality, and manifestly $O(h)$ for fixed $t_n$.