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  "body": "\n## Proof\n\nWe compare the partial sums of $e^{z+w}$ with products of partial sums, and bound the difference by the same comparison for $|z|$ and $|w|$.\n\n1. A sequence $(u_N)$ of complex numbers converges to $u$ when $|u_N - u| \\to 0$, with $|\\cdot|$ the [[modulus|0KW2X0TFBX]].\nSince $|\\operatorname{Re} v|, |\\operatorname{Im} v| \\le |v| \\le |\\operatorname{Re} v| + |\\operatorname{Im} v|$, this means that the real parts and the imaginary parts converge.\nThese inequalities follow by squaring, since $|v|^2 = (\\operatorname{Re} v)^2 + (\\operatorname{Im} v)^2$.\nLimits of complex sequences are unique, because limits of real sequences are.\n2. Fix $z, w \\in \\C$. For $j, k \\ge 0$ let $t_{j k} = \\dfrac{z^j w^k}{j! \\, k!}$.\nFor $N \\ge 0$ let $A_N = \\sum_{j=0}^{N} \\dfrac{z^j}{j!}$, $B_N = \\sum_{k=0}^{N} \\dfrac{w^k}{k!}$ and $D_N = \\sum_{n=0}^{N} \\dfrac{(z + w)^n}{n!}$.\nBy the definition of the [[complex exponential|054VXHMCA3]], $A_N \\to e^z$, $B_N \\to e^w$ and $D_N \\to e^{z+w}$.\n3. Since $\\C$ is a commutative ring by [[the complex numbers form a field|0FRSR10DWF]], the [[binomial theorem|0EHWPWGHB2]] applies.\nUsing $\\binom{n}{j} / n! = 1 / \\bigl(j! (n - j)!\\bigr)$, it gives $\\dfrac{(z + w)^n}{n!} = \\sum_{j + k = n} t_{j k}$.\n4. Let $T_N$ be the set of pairs $(j, k)$ of natural numbers with $j + k \\le N$, and $Q_N$ the set of pairs with $j \\le N$ and $k \\le N$.\nBy step 3, $D_N = \\sum_{(j,k) \\in T_N} t_{j k}$. By distributivity, $A_N B_N = \\sum_{(j,k) \\in Q_N} t_{j k}$.\n5. Since $T_N \\subseteq Q_N$, the difference is $\\Delta_N = A_N B_N - D_N = \\sum_{(j,k) \\in Q_N \\setminus T_N} t_{j k}$.\n6. Let $x = |z|$ and $y = |w|$. Define $a_N$, $b_N$, $d_N$ and $\\delta_N = a_N b_N - d_N$ by the same formulas with $x, y$ in place of $z, w$.\nBy steps 3 to 5 for $x$ and $y$, $\\delta_N = \\sum_{(j,k) \\in Q_N \\setminus T_N} \\dfrac{x^j y^k}{j! \\, k!}$.\n7. Since the [[modulus is multiplicative|0HYNZE940T]], $|t_{j k}| = \\dfrac{x^j y^k}{j! \\, k!}$.\nAlso $|\\operatorname{Re} t_{j k}| \\le |t_{j k}|$, since $(\\operatorname{Re} t)^2 \\le (\\operatorname{Re} t)^2 + (\\operatorname{Im} t)^2$.\n8. The real part of a sum is the sum of the real parts. By the [[triangle inequality|03HF5PMY5T]] for finite sums of reals and step 7, $|\\operatorname{Re} \\Delta_N| \\le \\delta_N$.\nIn the same way, $|\\operatorname{Im} \\Delta_N| \\le \\delta_N$.\n9. The series $\\sum x^j / j!$ and $\\sum y^k / k!$ converge to the real numbers $e^x$ and $e^y$. Their terms are nonnegative, so they converge absolutely.\n10. By the [[limit of a product|0734RSVF2V]], $a_N b_N \\to e^x e^y$.\n11. The [[Cauchy product|0B1K9PVWE9]] of these two series has $n$th term $\\sum_{j + k = n} x^j y^k / (j! \\, k!)$, so its partial sums are the $d_N$.\nBy the theorem on the [[Cauchy product of absolutely convergent series|0AN28XQ45Z]], $d_N \\to e^x e^y$.\n12. By steps 10 and 11 and the [[limit of a sum|0RAM4E07NR]], $\\delta_N \\to 0$.\n13. By step 8, $-\\delta_N \\le \\operatorname{Re} \\Delta_N \\le \\delta_N$. The [[squeeze theorem|0XMF00T4XQ]] gives $\\operatorname{Re} \\Delta_N \\to 0$. Likewise $\\operatorname{Im} \\Delta_N \\to 0$, so $\\Delta_N \\to 0$.\n14. We have $\\operatorname{Re}(A_N B_N) = \\operatorname{Re} A_N \\operatorname{Re} B_N - \\operatorname{Im} A_N \\operatorname{Im} B_N$, and a similar formula for the imaginary part.\nBy the limit rules for real products and sums, $A_N B_N \\to e^z e^w$.\n15. By steps 13 and 14, $D_N = A_N B_N - \\Delta_N \\to e^z e^w$.\n16. By step 2, also $D_N \\to e^{z+w}$. Limits are unique, so $e^{z + w} = e^z e^w$.\n\n## Idea\n\nMultiplying two partial sums gives all terms $z^j w^k / (j! k!)$ in a square, while the binomial theorem shows that $e^{z+w}$ only collects a triangle.\nThe leftover corner is bounded by the same corner for $|z|$ and $|w|$.\nFor nonnegative reals the corner tends to $0$, by the theorem on Cauchy products.\n",
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