First, notice that $a!+b!+c!>1!+1!+1!\implies 2^k>3\implies k\geq 2$ For $k=2\implies 2^2= a!+b!+c!\implies (k,a,b,c)= (2, 2, 1, 1)$ For $k=3\implies 2^3= a!+b!+c!\implies (k,a,b,c)= (3, 3, 1, 1)$ For $k=4\implies 2^4= a!+b!+c!\implies 16= a!+b!+c!$. But now, we see that: $\begin{cases} (i) (a,b,c)= (3,3,3)\implies 3!+3!+3!= 18\\ (ii) (a,b,c)= (3, 3, 2)\implies 3!+3!+2!= 14\end{cases}$ As we don't have any case between (i) and (ii), we don't have solutions for $k=4$. For $k=5\implies 2^5= a!+b!+c!\implies (k, a, b, c)= (5, 4, 3, 2)$ For $k=6\implies 2^6= a!+b!+c!\implies 64= a!+b!+c!.$ But now, we see that: $\begin{cases}(iii) (a,b,c)= (4, 4, 4)\implies 4!+4!+4!= 24+24+24= 72\\(iv) (a, b, c)= (4, 4, 3)\implies 4!+4!+3!= 24+24+6= 48+6= 54\end{cases}$. Similarly, as we did in the case for $k=4$, we don't have any case between (iii) and (iv), so we don't have solutions for $k=6$. For $k=7\implies 2^7= a!+b!+c!\implies (k, a, b, c)= (7, 5, 3, 2)$ Lemma: $\nexists (k,a,b,c)$ for $k\geq 8$ such that $2^k= a!+b!+c!$ and $a\geq b\geq c$ Proof: If $a\geq b\geq c>2\implies \begin{cases} 3\mid a!\\3\mid b!\\ 3\mid c!\end{cases}\implies 3\mid a!+b!+c!\implies 3\mid 2^k, k\geq 8$. That's impossible, because $2^k$ only has factors 2, and no factor 3. So, $a\geq b\geq c\leq 2$. Notice that $a\geq 5$, because $5!<2^8= 256$. Now, we only have to test 4 cases: Case 1: If $a\geq b=c=1$ $\implies a!+1!+1!= 2^k\implies a!= 2^k-2\implies a!= 2(2^{k-1}-1)$. As $a\geq 5!= 5\cdot 4\cdot 3\cdot 2\cdot 1\implies a\geq 5\cdot 3\cdot 4\cdot 2\implies 4\mid a!= 2(2^{k-1}-1)$. So, $\cancelto{2}{4}\mid \cancel{2}(2^{k-1}-1)\implies 2\mid 2^{k-1}-1$, which is false. Case 2: If $a\geq b=c=2$ $\implies a!+2!+2!= 2^k\implies a!+4= 2^k\implies a!= 4(2^{k-2}-1)$. As $a\geq 5!= 5\cdot 4\cdot 3\cdot 2\cdot 1\implies a\geq 5\cdot 3\cdot 8\implies 8\mid a!$, because these factors (2 and 4 will stay as we increase a). So, $\cancelto{2}{8}\mid a!= \cancel{4}(2^{k-1}-1)\implies 2\mid 2^{k-1}-1$, which is false. Case 3: If $a\geq b>2\geq c=1$ $\implies a!+b!+1!= 2^k\implies a!+b!= 2^k-1$. As $a, b>2\implies \begin{cases} 2\mid a!\\ 2\mid b!\end{cases}\implies 2\mid a!+b!= 2^k-1\implies 2\mid 2^k -1$, which is false. Case 4: If $a\geq b=2\geq c=1$ $\implies a!+2!+1!= 2^k \implies a!+3= 2^k\implies a!= 2^k-3$. As $a\geq 5\implies 3\mid a!= 2^k-3\implies 3\mid 2^k$, which is false, because $2^k$ only has factors 2, and none of factor 3. Therefore, doesn't exist solutions $(k,a,b,c)$ for $k\geq 8$. Now, we conclude with our 4 solutions $(k, a, b, c)= (2, 2, 1, 1); (3, 3, 1, 1); (5, 4, 3, 2); (7, 5, 3, 2)$